A manufacturing machine has a 40% defect rate. If 131 items are chosen at random, answer the following. a) Which is the correct wording for the random variable? rv X = the number of 131 randomly selected items that are defective Correct b) Pick the correct symbol: n Correct = 131 c) Pick the correct symbol: p Correct = 0.4

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.

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]

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

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.

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]

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

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

Answer:

I need more explanation is there more to the question?

not an answer but is this what your doing?

Answer:its 15.7

Step-by-step explanation:

Answer:

15.7

Step-by-step explanation:

Answer:

The car's average 18 number of miles per gallon

Step-by-step explanation:

The car has different efficiencies depending if its used on highway or city streets.

The efficiency on a highway is:

[tex]E_h=\dfrac{350}{15}\;\text{miles/gal}=23.33\;\text{miles/gal}[/tex]

The efficiency for city driving is:

[tex]E_c=\dfrac{280}{18}\;\text{miles/gal}=15.56\;\text{miles/gal}[/tex]

We now that the car uses twice as many gallons for city driving as for highway driving. This means that, for every gallon consumed, 2/3 are for city driving and 1/3 are for highway driving.

Then, we can calculate how many miles makes the car on average as:

[tex]E=\dfrac{1}{3}E_h+\dfrac{2}{3}E_c\\\\\\E=\dfrac{1}{3}\cdot 23.33+\dfrac{2}{3}\cdot15.56=7.78+10.37=18.15\;\text{miles/gal}[/tex]

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

:))

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

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.

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Answer:

0.05

Step-by-step explanation:

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:

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;

Which gives;

[tex]V = \pi \times 3^2 \times 10 - \frac{2}{3} \times \pi \times 3^3 \approx \mathbf{ 226}[/tex]

(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

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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]

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.

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

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.

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:

A 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

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Answer:

solution

Step-by-step explanation:

x=5

y=4

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

Answer:

Step-by-step explanation:

Idk

Answer:

the triangles are not similar.

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

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

Answer:

$301.23

Step-by-step explanation:

We have that the function of wealth is U (w) = w ^ (1/2)

So, since what you have at the start is 100, we replace:

U (w) = 100 ^ (1/2)

U = 10

Now we have two cases:

the first one we win, then the winnings would be 100 minus the cost of the lottery, that is 36 and to that add G of the prize:

100 - 36 + G = 64 + G

In the second case, where we lose, the subtraction of 100 that we have minus the cost of the lottery would be equal 36

100 - 36 = 64

Therefore, we have to win with an 18% probability, therefore losing would be 82% (100% - 18%)

0.18 * (64 + G) ^ (1/2) + 0.82 * 64 ^ (1/2)

solving:

0.18 * (64 + G) ^ (1/2) + 6.56

Now this is equal to U which is equal to 10:

10 = 0.18 * (64 + G) ^ (1/2) + 6.56

(10 - 6.56) /0.18 = (64 + G) ^ (1/2)

(64 + G) ^ (1/2) = 19.11

(64 + G) = 365.23

G = 365.23 - 64

G = 301.23

Therefore, the smallest G prize size that the lottery will be willing to buy is $ 301.23

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.

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]

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

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]