Answer:
GH¢. 18098.46
Step-by-step explanation:
Let the first investment giving 12% interest per annum be Bank A
Let the 2nd investment giving 10% per annum be bank B
Let the first amount invested be
GH¢. X and let the second amount invested be GH¢. X + 580
Thus; In bank A;
Principal amount in first = GH¢. x
rate = 12 %
time = 1 year
Formula for simple interest = PRT/100
Where P is principal, R is rate and T is time.
So, interest in his investment = 12X/100 = 0.12X
while in bank B;
principal amount = GH¢. X + 580
rate = 14%
time = 1 yr
So, interest in his investment = [(X + 580) × 14]/100
= 0.14(X + 580)
So, total accumulated interest is;
0.12X + 0.14(X + 580) = 0.12X + 0.14X + 81.2 = 0.26X + 81.2
Now, we are given accumulated interest = GH¢. 2,358.60
Thus;
2358.60 = (0.26X + 81.2)
2358.6 - 81.2 = 0.26X
X = 2277.4/0.26
X = 8759.23
So,
first amount invested = GH¢. 8759.23
Second amount invested = GH¢. 8759.23 + GH¢. 580 = GH¢. 9339.23
Total amount invested = GH¢. 8759.23 + GH¢. 9339.23 = GH¢. 18098.46
If f(x) = 4 – x2 and g(x) = 6x, which expression is equivalent to (9-1(3)?
Answer:
( g − f ) ( 3 ) = 23
Step-by-step explanation:
(g-f)(x)=g(x)-f(x)
=6x-(4-X(2))
=x(2)+6x-4
to evaluate (g-f) (#) substitute x=3 into (g-f)(x)
(g-f)=(9)+(6 x 3) -4=23
A car dealership decreased the price of a certain car by 4% . The original price was $43,600 . write the new price in terms of the original price.
Answer: The new price of the car is $41856
Step-by-step explanation:
So we know the the original price as 43,600 which is 100% and is being dropped by 4% so you would have to subtract 4% from a 100% and multiply it by the original price.
100% - 4% = 96%
Now 96% of the original price is the new price.
96% * 43,600= ?
0.96 * 43,600 = 41856
what is the value of x in the equation 2x+3y=36 when y=6
Answer:
9
Step-by-step explanation:
[tex]2x+3y=36\\\\2x+3(6)=36\\\\2x+18=36\\\\2x=18\\\\x=9[/tex]
Hope this helps!
Answer:
X= 9
Step-by-step explanation:
2x+3y=36
2x+3(6)=36
2x+18=36
-18 -18
2x=18
----------
2
x=9
Andrei wants to fill a glass tank with marbles, and then fill the remaining space with water. WWW represents the volume of water Andrei uses (in liters) if he uses nnn marbles. W=32-0.05nW=32−0.05nW, equals, 32, minus, 0, point, 05, n What is the glass tank's volume?
Before Andrei adds the marbles to the glass tank, the glass tank was empty. This means that the volume of the empty tank is when n = 0 and the volume is 32 liters.
Given that:
[tex]W = 32 - 0.05n[/tex]
A linear function is represented as:
[tex]y = b + mx[/tex]
Where
[tex]b \to[/tex] y intercept
Literally, the y intercept is the initial value of the function.
In this function, the y intercept means the initial volume of the glass tank before filling it with marbles.
Compare [tex]y = b + mx[/tex] and [tex]W = 32 - 0.05n[/tex]
[tex]b = 32[/tex]
This means that the volume of the glass tank is 32 liters.
Read more about linear functions at:
https://brainly.com/question/21107621
Answer:
0.05
Step-by-step explanation:
One angle of a right triangle measures 51 degrees. What is the measure of the other small angle?
Answer:
39 degrees
Step-by-step explanation:
Given
triangle is right angled i.e one angle is 90 degrees
other angle is 51 degrees.
let the third angle be x degrees
we know that sum of angles of any triangle is 180 degrees
thus,
90 + 51+ x = 180
=> 141 + x = 180
=> x = 180 - 141 = 39.
Thus, measure of other small angle is 39 degrees.
Answer:
Step-by-step explanation:
m∠A+m∠B+m∠C = 180
90+51+x1= 180
41+x=180
x=39
What are the next two numbers in the pattern of numbers;
45, 15, 44, 17, 40, 20, 31, 25, …
Answer:
Next two numbers are 15 and 32 respectively.
Step-by-step explanation:
The given pattern is
45, 15, 44, 17, 40, 20, 31, 25, …
Here, we have two patterns.
Odd places : 45, 44, 40, 31,...
Even places : 15, 17, 20, 25,...
In series of odd places, we need to subtract square of integers.
[tex]45-(1)^2=45-1=44[/tex]
[tex]44-(2)^2=44-4=40[/tex]
[tex]40-(3)^2=40-9=31[/tex]
So, 9th term of given pattern is
[tex]31-(4)^2=31-16=15[/tex]
In series of even places, we need to add prime numbers.
[tex]15-2=17[/tex]
[tex]17+3=20[/tex]
[tex]20+5=25[/tex]
So, 10th term of given pattern is
[tex]25+7=32[/tex]
Therefore, the next two numbers in the pattern of numbers are 15 and 32 respectively.
Two forces of magnitudes 16 and 20 pounds are acting on an object. The bearings of the forces are N75E and S20E (that is, 75∘ east of north and 20∘ east of south), respectively. How many degrees east of south is the resultant force? (Round to two decimal places and do not enter the ∘ symbol.)
Answer:
20.58
Step-by-step explanation:
Force A = 16 pounds
direction = N75E = 75° in the first quadrant
Force B = 20 pounds
direction = S20E = 180° - 20° = 160° in the fourth quadrants
we resolve the forces into their x and y resultant forces.
For the y axis forces,
-(16 x sin 75°) + (20 x sin 160°) = Fy
-15.45 + 6.84 = Fy
Fy = -8.61 N
For the x axis forces'
(16 x cos 75°) + (20 x cos 160°) = Fx
-4.14 - 18.79 = Fx
Fx = -22.93 N
Resultant force = [tex]\sqrt{(Fx)^{2} + (Fy)^{2} }[/tex] = [tex]\sqrt{(-8.61)^{2} + (-22.93)^{2} }[/tex]
Resultant force = 24.49 N
angle made by the resultant force is,
∅ = [tex]tan^{-1}[/tex] [tex]\frac{Fy}{Fx}}[/tex]
∅ = [tex]tan^{-1}[/tex] [tex]\frac{-8.61}{-22.93}}[/tex]
∅ = [tex]tan^{-1}[/tex] 0.3755
∅ = 20.58
what is the value of x?
Answer:
x = 5
Step-by-step explanation:
52 = y since they are the base angles of an isosceles triangle and the base angles are equal
The sum of the angles of a triangle are 180
52+y+14x+6 =180
Substitute for y
52+52+14x+6 = 180
Combine like terms
110 + 14x = 180
Subtract 110 from each side
110+14x-110 = 180-110
14x =70
Divide by 14
14x/14 = 70/14
x =5
An engineering consulting firm wantedto evaluate a rivet process by measuring the formed diameter. The following data represent the diameters (in hundredths of an inch) for a random sample of 24 rivet heads:
6.81 - 6.79 - 6.69 - 6.59 - 6.65 - 6.60 - 6.74 - 6.70 - 6.76
6.84 - 6.81 - 6.71 - 6.66 - 6.76 - 6.76 - 6.77 - 6.72 - 6.68
7.71 - 6.79 - 6.72 - 6.72 - 6.72 - 6.79 - 6.83
a) Set up a 95% confidence interval estimate of the average diameter of rivet heads (in hundredths of an inch).
b) Set up a 95% confidence interval estimate of the standard deviation of the diameter of rivet heads (in hundredths of an inch)
Answer:
Step-by-step explanation:
6.81 - 6.79 - 6.69 - 6.59 - 6.65 - 6.60 - 6.74 - 6.70 - 6.76
6.84 - 6.81 - 6.71 - 6.66 - 6.76 - 6.76 - 6.77 - 6.72 - 6.68
7.71 - 6.79 - 6.72 - 6.72 - 6.72 - 6.79 - 6.83
[tex]\bar x =6.77[/tex]
S.D = 0.21
[tex]I=6.77\pmt\times\frac{s}{\sqrt{n} }[/tex]
df = 24
α = 0.05
t = 2.064
[tex]I=6.77\pm2.064\times\frac{0.21}{\sqrt{25} } \\\\=6.77\pm0.087\\\\=[6.683,6.857][/tex]
b)
[tex]\sqrt{\frac{(1-n)s^2}{X^2_{\alpha /2} } < \mu <\sqrt{\frac{(1-n)s^2}{X^2_{1-\alpha/2} } }[/tex]
[tex]\sqrt{\frac{24 \times 0.21^2}{39.364} } < \mu <\sqrt{\frac{24 \times 0.21^2}{12.401} } \\\\=0.1640<\mu<0.2921[/tex]
the price of a CD that sells for 21% more than
the amount (m) needed to manufacture the CD
Answer:
I need more explanation is there more to the question?
not an answer but is this what your doing?
What is the missing number in the pattern? Please Help. Been stuck on this for hours.
Answer:
8
Step-by-step explanation:
The other patterns go with the two factors on top (2 x 3 = 6 and 3 x 3 = 9).
So, 2 x 4 = 8
A design engineer wants to construct a sample mean chart for controlling the service life of a halogen headlamp his company produces. He knows from numerous previous samples that when this service life is in control it is normally distributed with a mean of 500 hours and a standard deviation of 20 hours. On three recent production batches, he tested service life on random samples of four headlamps, with these results: Sample Service Life (hours) 1 495 500 505 500 2 525 515 505 515 3 470 480 460 470 What is the mean of the sampling distribution of sample means when the service life is in control
Answer:
[tex]$ \text {Sample mean} = \bar{x} = \mu = 500 \: hours $[/tex]
Step-by-step explanation:
What is Normal Distribution?
We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.
For the given scenario, it is known from numerous previous samples that when this service life is in control it is normally distributed with a mean of 500 hours and a standard deviation of 20 hours.
On three recent production batches, he tested service life on random samples of four headlamps.
We are asked to find the mean of the sampling distribution of sample means when the service life is in control.
Since we know that the population is normally distributed and a random sample is taken from the population then the mean of the sampling distribution of sample means would be equal to the population mean that is 500 hours.
[tex]$ \text {Sample mean} = \bar{x} = \mu = 500 \: hours $[/tex]
Whereas the standard deviation of the sampling distribution of sample means would be
[tex]\text {standard deviation} = s = \frac{\sigma}{\sqrt{n} } \\\\[/tex]
Where n is the sample size and σ is the population standard deviation.
[tex]\text {standard deviation} = s = \frac{20}{\sqrt{4} } \\\\ \text {standard deviation} = s = \frac{20}{2 } \\\\ \text {standard deviation} = s = 10 \: hours \\\\[/tex]
A machine has four components, A, B, C, and D, set up in such a manner that all four parts must work for the machine to work properly. Assume the probability of one part working does not depend on the functionality of any of the other parts. Also assume that the probabilities of the individual parts working are P(A) = P(B) = 0.93, P(C) = 0.95, and P(D) = 0.92. Find the probability that the machine works properly.A) 0.8128B) 0.2441C) 0.8217D) 0.7559
Answer:
D) 0.7559
Step-by-step explanation:
Independent events:
If two events, A and B are independent:
[tex]P(A \cap B) = P(A)*P(B)[/tex]
In this question:
Four independent events, A, B, C and D.
So
[tex]P(A \cap B \cap C \cap D) = P(A)*P(B)*P(C)*P(D)[/tex]
Find the probability that the machine works properly.
This is the probability that all components work properly.
[tex]P(A \cap B \cap C \cap D) = P(A)*P(B)*P(C)*P(D) = 0.93*0.93*0.95*0.92 = 0.7559[/tex]
So the correct answer is:
D) 0.7559
In monitoring lead in the air after the explosion at the battery factory, it is found that the amounts of lead over a 6 day period had a standard error of 1.93. Find the margin of error that corresponds to a 95% confidence interval. (Round to 2 decimal places) 4.56
Answer:
1.54
Margin of error M.E = 1.54
Step-by-step explanation:
Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.
The confidence interval of a statistical data can be written as.
x+/-zr/√n
x+/-M.E
Where M.E = margin of error
M.E = zr/√n
Given that
Standard deviation r = 1.93
Number of samples n = 6
Confidence interval = 95%
z(at 95% confidence) = 1.96
Substituting the values we have;
M.E = (1.96×1.93/√6) = 1.544321633166
M.E = 1.54 (to 2 decimal place)
Margin of error M.E = 1.54
One number is 3 more than 2 times the other, and their sum is 27. Find the numbers.
If x represents the smaller number, then the larger number is
3x + 2
2x + 3
21x + 3)
Answer:
Option 2 is correct
Step-by-step explanation:
One number is 2 times another number plus 3. Their sum is 21.
"One number is 2 times another number plus 3" translated to
x = smaller number = another number
It is also given that: Their sum is 21.
Combine like terms:
3x+3 = 21
Answer:
I do questions like these everyday so I have too much experience. Let me explain step by step for you.
Brainliest?
First lets set 2 variables x and y
Lets make 2 equations.
x=3+2*y
Thats because it says 'x' is 3 more (+) than 2 times(*) 'y'
Now lets set second, we know both of them add up to 27.
x+y = 27
Since we know what x is equal to (look above equation)
We can replace it.
x is replaced with 3+2*y
3+2y+y = 27
3+4y = 27
Simplify 27-3 = 24
24/4 = 6
Now lets plug in for x
3+2*6 = 15
15 - x
6 - y
:))
What is the approximate length of minor arc LM? Round to
the nearest tenth of a centimeter.
12.4 centimeters
15.7 centimeters
31.4 centimeters
36.7 centimeters
Answer:its 15.7
Step-by-step explanation:
Answer:
15.7
Step-by-step explanation:
what is the value of x?
Answer:
solution
Step-by-step explanation:
x=5
y=4
Zia is building a plastic model rocket that has the combined shape of a cone and a cylinder as shown. additionally, the cylinder has a hemisphere hollowed out of its bottom. the plastic for the cone weighs 1.4 grams per cubic centimeter and the plastic for the cylinder weights only 0.8 grams per cubic centimeter.
(a) the volume of plastic that remains in the cylinder after it has been hollowed out to the nearest cubic centimeter.
(b) what has a greater total weight, the plastic that makes up the cone or the plastic that makes up the cylinder after it has been hollowed out?
Answer:
226 cm^3
The mass of plastic used to make cylinder is greater
Step-by-step explanation:
Given:-
- The density of cone material, ρc = 1.4 g / cm^3
- The density of cylinder material, ρl = 0.8 g / cm^3
Solution:-
- To determine the volume of plastic that remains in the cylinder after gouging out a hemispherical amount of material.
- We will first consider a solid cylinder with length ( L = 10 cm ) and diameter ( d = 6 cm ). The volume of a cylinder is expressed as follows:
[tex]V_L =\pi \frac{d^2}{4} * L[/tex]
- Determine the volume of complete cylindrical body as follows:
[tex]V_L = \pi \frac{(6)^2}{4} * 10\\\\V_L = 90\pi cm^3\\[/tex]
- Where the volume of hemisphere with diameter ( d = 6 cm ) is given by:
[tex]V_h = \frac{\pi }{12}*d^3[/tex]
- Determine the volume of hemisphere gouged out as follows:
[tex]V_h = \frac{\pi }{12}*6^3\\\\V_h = 18\pi cm^3[/tex]
- Apply the principle of super-position and subtract the volume of hemisphere from the cylinder as follows to the nearest ( cm^3 ):
[tex]V = V_L - V_h\\\\V = 90\pi - 18\pi \\\\V = 226 cm^3[/tex]
Answer: The amount of volume that remains in the cylinder is 226 cm^3
- The volume of cone with base diameter ( d = 6 cm ) and height ( h = 5 cm ) is expressed as follows:
[tex]V_c = \frac{\pi }{12} *d^2 * h[/tex]
- Determine the volume of cone:
[tex]V_c = \frac{\pi }{12} *6^2 * 5\\\\V_c = 15\pi cm^3[/tex]
- The mass of plastic for the cylinder and the cone can be evaluated using their respective densities and volumes as follows:
[tex]m_i = p_i * V_i[/tex]
- The mass of plastic used to make the cylinder ( after removing hemispherical amount ) is:
[tex]m_L = p_L * V\\\\m_L = 0.8 * 226\\\\m_L = 180.8 g[/tex]
- Similarly the mass of plastic used to make the cone would be:
[tex]m_c = p_c * V_c\\\\m_c = 1.4 * 15\pi \\\\m_c = 65.973 g[/tex]
Answer: The total weight of the cylinder ( m_l = 180.8 g ) is greater than the total weight of the cone ( m_c = 66 g ).
The volume of the remaining plastic in the cylinder is large, which
makes the weight much larger than the weight of the cone.
Responses:
(a) Volume of the remaining plastic in the cylinder is 226 cm³(b) The weight of the cylinder is greater than the weight of the cone.How can the weight and volume be evaluated?Density of the plastic for the cone = 1.4 g/cm³
Density of the plastic used for the cylinder = 0.8 g/cm³
From a similar question, we have;
Height of the cylinder = 10 cm
Diameter of the cylinder = 6 cm
Height of the cone = 5 cm
(a) Radius of the cylinder, r = 6 cm ÷ 2 = 3 cm
Volume of a cylinder = π·r²·h
Volume of a hemisphere = [tex]\mathbf{\frac{2}{3}}[/tex] × π× r³
Volume of the cylinder after it has been hollowed out, V, is therefore;
[tex]V = \mathbf{\pi \times r^2 \times h - \frac{2}{3} \times \pi \times r^3}[/tex]Which gives;
[tex]V = \pi \times 3^2 \times 10 - \frac{2}{3} \times \pi \times 3^3 \approx \mathbf{ 226}[/tex]
Volume of the cylinder after it has been hollowed out, V ≈ 226 cm³(b) Volume of the cone = [tex]\mathbf{\frac{1}{3}}[/tex] × π × 3² × 5 ≈ 47.1
Mass of the cone = 47.1 cm³ × 1.4 g/cm³ ≈ 66 g
Mass of the hollowed cylinder ≈ 226 cm³ × 0.8 g/cm³ = 180.8 g
The mass and therefore, the weight of the plastic that makes up the hollowed cylinder is greater than the weight of the plastic that makes up the cone.Learn more about volume and density of solids here:
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which answer shows 9 x 10 ^ -5 written in standard form ?
A -0.000009
B -0.00009
C 0.0009
D 0.00009
Answer:
D 0.00009
Step-by-step explanation:
9 × 10^-5 = 9 × 1/10^5 = 9 × 1/100,000
= 9 × 0.00001
= 0.00009
_____
Comment on place value
The exponent of 10 associated with the place value in a decimal number increases from 0 to the left of the decimal point, and decreases from -1 to the right of the decimal point:
100. = 10²
10. = 10¹
1. = 10⁰
0.1 = 10⁻¹
0.01 = 10⁻²
0.001 = 10⁻³
0.0001 = 10⁻⁴
0.00001 = 10⁻⁵
This simple realization can help you immensely with scientific notation.
What weighs more a pound of feathers or a pound of pennies
What percentage of the total number of microstates are in one of the three most likely macro states of 100 coins being tossed (49 heads and 51 tails, 50 heads and 50 tails, or 51 heads and 49 tails)
Answer:
Step-by-step explanation:
Idk
The HCF of two numbers is 11, and their L.C.M is 368. If one number is 64, then the other number is ….
Answer:
63.45
Step-by-step explanation:
First, it should be noted that the question is incorrect because 64 can't be divided by 11 but assuming that the question is correct, the solution is as follows
Given
LCM = 368
HCF = 11
One number = 64
Required
The other number
Let both numbers be represented by m and n, such that
[tex]m = 64[/tex]
From laws of HCF and LCM
The product of both numbers = Product of HCF and LCM
i.e.
[tex]m * n = HCF * LCM[/tex]
By substituting 68 for m; 368 for LCM and 11 for HCF
[tex]m * n = HCF * LCM[/tex] becomes
[tex]64 * n = 368 * 11[/tex]
[tex]64n = 4048[/tex]
Divide both sides by 64
[tex]\frac{64n}{64} = \frac{4048}{64}[/tex]
[tex]n = \frac{4048}{64}[/tex]
[tex]n = 63.25[/tex]
i need help in homework no guess
Answer:
No
Step-by-step explanation:
Use the vertical line test. If the line intercepts more than one point, it is not a function. Since there are two points where the value of 'x' is two, the line will pass both points. The graph is not a function.
A production facility employs 10 workers on the day shift, 8 workers on the swing shift, and 6 workers on the graveyard shift. A quality control consultant is to select 4 of these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of 4 workers has the same chance of being selected as does any other group (drawing 4 slips without replacement from among 24).
(a) How many selections result in all 4 workers coming from the day shift? What is the probability that all 4 selected workers will be from the day shift? (Round your answer to four decimal places.)
(b) What is the probability that all 4 selected workers will be from the same shift? (Round your answer to four decimal places.)
(c) What is the probability that at least two different shifts will be represented among the selected workers? (Round your answer to four decimal places.)
(d) What is the probability that at least one of the shifts will be unrepresented in the sample of workers? (Round your answer to four decimal places.)
The probability that all 4 selected workers will be from the day shift is, = 0.0198
The probability that all 4 selected workers will be from the same shift is = 0.0278
The probability that at least two different shifts will be represented among the selected workers is = 0.9722
The probability that at least one of the shifts will be unrepresented in the sample of workers is P(A∪B∪C) = 0.5257
To solve this question properly, we will need to make use of the concept of combination along with set theory.
What is Combination?In mathematical concept, Combination is the grouping of subsets from a set without taking the order of selection into consideration.
The formula for calculating combination can be expressed as:
[tex]\mathbf{(^n _r) =\dfrac{n!}{r!(n-r)! }}[/tex]
From the parameters given:
Workers employed on the day shift = 10Workers on swing shift = 8Workers on graveyard shift = 6A quality control consultant is to select 4 of these workers for in-depth interviews:
Using the expression for calculating combination:
(a)
The number of selections results in all 4 workers coming from the day shift is :
[tex]\mathbf{(^n _r) = (^{10} _4)}[/tex]
[tex]\mathbf{=\dfrac{(10!)}{4!(10-4)!}}[/tex]
= 210
The probability that all 5 selected workers will be from the day shift is,
[tex]\begin{array}{c}\\P\left( {{\rm{all \ 4 \ selected \ workers\ will \ be \ from \ the \ day \ shift}}} \right) = \dfrac{{\left( \begin{array}{l}\\10\\\\4\\\end{array} \right)}}{{\left( \begin{array}{l}\\24\\\\4\\\end{array} \right)}}\\\end{array}[/tex]
[tex]\mathbf{= \dfrac{210}{10626}} \\ \\ \\ \mathbf{= 0.0198}[/tex]
(b) The probability that all 4 selected workers will be from the same shift is calculated as follows:
P( all 4 selected workers will be) [tex]\mathbf{= \dfrac{ \Big(^{10}_4\Big) }{\Big(^{24}_4\Big)}+\dfrac{ \Big(^{8}_4\Big) }{\Big(^{24}_4\Big)} + \dfrac{ \Big(^{6}_4\Big) }{\Big(^{24}_4\Big)}}[/tex]
where;
[tex]\mathbf{\Big(^{8}_4\Big) = \dfrac{8!}{4!(8-4)!} = 70}[/tex]
[tex]\mathbf{\Big(^{6}_4\Big) = \dfrac{6!}{4!(6-4)!} = 15}[/tex]
P( all 4 selected workers is:)
[tex]\mathbf{=\dfrac{210+70+15}{10626}}[/tex]
The probability that all 4 selected workers will be from the same shift is = 0.0278
(c)
The probability that at least two different shifts will be represented among the selected workers can be computed as:
[tex]= 1-\dfrac{ (^{10}_4) }{(^{24}_4)}+\dfrac{ (^{8}_4) }{(^{24}_4)} + \dfrac{ (^{6}_4) }{(^{24}_4)}[/tex]
[tex]=1 - \dfrac{210+70+15}{10626}[/tex]
= 1 - 0.0278
= 0.9722
The probability that at least two different shifts will be represented among the selected workers is = 0.9722
(d)
The probability that at least one of the shifts will be unrepresented in the sample of workers is:
[tex]P(AUBUC) = \dfrac{(^{6+8}_4)}{(^{24}_4)}+ \dfrac{(^{10+6}_4)}{(^{24}_4)}+ \dfrac{(^{10+8}_4)}{(^{24}_4)}- \dfrac{(^{6}_4)}{(^{24}_4)}-\dfrac{(^{8}_4)}{(^{24}_4)}-\dfrac{(^{10}_4)}{(^{24}_4)}+0[/tex]
[tex]P(AUBUC) = \dfrac{(^{14}_4)}{(^{24}_4)}+ \dfrac{(^{16}_4)}{(^{24}_4)}+ \dfrac{(^{18}_4)}{(^{24}_4)}- \dfrac{(^{6}_4)}{(^{24}_4)}-\dfrac{(^{8}_4)}{(^{24}_4)}-\dfrac{(^{10}_4)}{(^{24}_4)}+0[/tex]
[tex]P(AUBUC) = \dfrac{1001}{10626}+ \dfrac{1820}{10626}+ \dfrac{3060}{10626}-\dfrac{15}{10626}-\dfrac{70}{10626}-\dfrac{210}{10626} +0[/tex]
The probability that at least one of the shifts will be unrepresented in the sample of workers is P(A∪B∪C) = 0.5257
Learn more about combination and probability here:
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A public relations firm found that only 27% of voters in a certain state are satisfied with their U.S. senators. How large a sample of voters should be drawn so that the sample proportion of voters who are satisfied with their senators is approximately normally distributed?a) 38b) 14c) 10d) 48
Answer:
a) 38
Step-by-step explanation:
The normal distribution can be applied if:
[tex]np \geq 5[/tex] and [tex]n(1-p) \geq 5[/tex]
In this question:
[tex]p = 0.27[/tex]
Then
a) 38
n = 38.
Then
38*0.27 = 10.26
38*0.73 = 27.74
Satisfies. But is this the smallest sample of the options which satisfies.
b) 14
n = 14
Then
14*0.23 = 3.22
14*0.77 = 10.78
Does not satisfy
c) 10
Smaller than 14, which also does not satisfy, so 10 does not satisfy.
d) 48
Greater than 38, which already satisfies. So the answer is a)
A stone is thrown vertically into the air at an initial velocity of 79 ft/s. On a different planet, the height s (in feet) of the stone above the ground after t seconds is sequals79tminus3t squared and on Earth it is sequals79tminus16t squared. How much higher will the stone travel on the other planet than on Earth?
Answer:
[tex]13t^2[/tex] feet higher the stone will travel on the other plant than on Earth.
Step-by-step explanation:
Initial velocity of the stone thrown vertically = 79 ft/s
It is given that:
Height attained on a different planet with time [tex]t[/tex]:
[tex]s_p = 79t -3t^2[/tex]
Height attained on Earth with time [tex]t[/tex]:
[tex]s_e = 79t -16t^2[/tex]
If we have a look at the values of [tex]s_p\text{ and }s_e[/tex], it can be clearly seen that the part [tex]79t[/tex] is common in both of them and some values are subtracted from it.
The values subtracted are [tex]3t^2\text{ and } 16t^2[/tex] respectively.
[tex]t^2[/tex] can never be negative because it is time value.
So, coefficient of [tex]t^2[/tex] will decide which is larger value that is subtracted from the common part i.e. [tex]79t[/tex].
Clearly, [tex]3t^2\text{ and } 16t^2[/tex] have [tex]16t^2[/tex] are the larger value, hence [tex]s_e < s_p[/tex].
So, difference between the height obtained:
[tex]s_p - s_e = 79t - 3t^2 - (79t - 16t^2)\\\Rightarrow 79t -3t^2 - 79t + 16t^2\\\Rightarrow 13t^2[/tex]
So, [tex]13t^2[/tex] feet higher the stone will travel on the other plant than on Earth.
intext:"A shipment of 50 inexpensive digital watches, including 6 that are defective, is sent to a department store. The receiving department selects 10 at random for testing and rejects the whole shipment if 1 or more in the sample are found defective. What is the probability that the shipment will be rejected?"
Answer:
0.7125
Step-by-step explanation:
The binomial distribution with parameters n and p is the discrete probability distribution of the number of successes (with probability p) in a sequence of n independent events.
The probability of getting exactly x successes in n independent Bernoulli trials = [tex]n_{C_{x}}(p)^x(1-p)^{n-x}[/tex]
Total number of watches in the shipment = 50
Number of defective watches = 6
Number of selected watches = 10
Let X denotes the number of defective digital watches such that the random variable X follows a binomial distribution with parameters n and p.
So,
Probability of defective watches = [tex]\frac{X}{n}=\frac{6}{50}=0.12[/tex]
Take n = 10 and p = 0.12
Probability that the shipment will be rejected = [tex]P(X\geq 1)=1-P(X=0)[/tex]
[tex]=1-n_{C_{x}}(p)^x(1-p)^{n-x}\\=1-10_{C_{0}}(0.12)^0(1-0.12)^{10-0}[/tex]
Use [tex]n_{C_{x}}=\frac{n!}{x!(n-x)!}[/tex]
So,
Probability that the shipment will be rejected = [tex]=1-\left ( \frac{10!}{0!(10-0)!} \right )(0.88)^{10}[/tex]
[tex]=1-(0.88)^{10}\\=1-0.2785\\=0.7125[/tex]
Use slope-intercept form to write the equation of a line
that has a slope of -3 and passes through the point
(1,-5).
Use the drop-down menus to select the proper value
for each variable that is substituted into the slope-
intercept equation
y =
X
DPM
m =
Answer:
y=-3x-2
Step-by-step explanation:
There is enough information to make a point-slope form equation that which we can convert into slope-intercept form.
Point-slope form is: [tex]y-y_1=m(x-x_1)[/tex]
We are given the slope of -3 and the point of (1,-5).
[tex]y-y_1=m(x-x_1)\rightarrow y+5=-3(x-1)[/tex]
Convert into Slope-Intercept Form:
[tex]y+5=-3(x-1)\\y+5-5=-3(x-1)-5\\\boxed{y=-3x-2}[/tex]
What is the next number in the sequence: 3, 8, 12, 48, 29, __
Answer:
144
Step-by-step explanation:
Answer:
116
Step-by-step explanation:
3x4=12
12x4=48
8x4=32
32-3=29
29x4=116
Hope it's clear
The amount of coffee that people drink per day is normally distributed with a mean of 17 ounces and a standard deviation of 4 ounces. 15 randomly selected people are surveyed. Round all answers to 4 decimal places where possible.
a) What is the distribution of XX? XX ~ N(,)
b) What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)
c) What is the probability that one randomly selected person drinks between 15.5 and 18 ounces of coffee per day?
d) For the 15 people, find the probability that the average coffee consumption is between 15.5 and 18 ounces of coffee per day.
e) For part d), is the assumption that the distribution is normal necessary? YesNo
f) Find the IQR for the average of 15 coffee drinkers.
Q1 = ounces
Q3 = ounces
IQR: ounces
Answer:
Step-by-step explanation:
(a)
The distribution of X is Normal Distribution with mean [tex]= \mu =17[/tex] and Variance [tex]= \sigma^{2} = 16 \ i.e., X \sim N (17, 16),[/tex]
(b)
The distribution of [tex]\bar{x}[/tex] is Normal Distribution with mean [tex]= \mu =17[/tex] and Variance = [tex]\sigma^{2}/n = 16/15= 1.0667[/tex].i.e., [tex]\bar{x}\sim N(17,1.0667)[/tex]
c)
To find P(15.5 < X < 18):
Case 1: For X from 15.5 to mid value:
Z = (15.5 - 17)/4 = - 0.375
Table of Area Under Standard Normal Curve gives area = 0.1480
Case 2: For X from mid value to 18:
Z = (18 - 17)/4 = 0.25
Table of Area Under Standard Normal Curve gives area = 0.0987
So,
P(15.5 < X< 18) = 0.1480 +0.0987 = 0.2467
So,
Answer is:
0.2467
(d)
[tex]SE = \sigma/\sqrt{n}\\\\= 4/\sqrt{15}[/tex]
= 1.0328
To find [tex]P(15.5 < \bar{x}< 18):[/tex]
Case 1: For [tex]\bar{x}[/tex] from 15.5 to mid value:
Z = (15.5 - 17)/1.0328 = - 1.4524
Table of Area Under Standard Normal Curve gives area = 0.4265
Case 2: For X from mid value to 18:
Z = (18 - 17)/1.0328 = 0.9682
Table of Area Under Standard Normal Curve gives area = 0.3340
So,
[tex]P(15.5 < \bar{x}< 18) = 0.4265 + 0.3340 = 0.7605[/tex]
So,
Answer is:
0.7605
(e)
Correct option:
No
because Population SD is provided.
(f)
(i)
Q1 is given by:
[tex]- 0.6745 = (\bar{x} - 17)/1.0328[/tex]
So,
X = 17 - (0.6745 * 1.0328) = 17 - 0.6966 = 16.3034
So,
Q1 = 16.3034
(ii)
Q3 is given by:
[tex]0.6745 = (\bar{x} - 17)/1.0328[/tex]
So,
X = 17 + (0.6745 * 1.0328) = 17 + 0.6966 = 17.6966
So,
Q3= 17.6966
(iii)
IQR = Q3 - Q1 = 17.6966 - 16.3034 = 1.3932
So
Answers are:
Q1 = 16.3034 ounces
Q3 = 17.6966 Ounces
IQR = 1.3932 Ounces