Answer:
[tex]F=\frac{s^2_1}{s^2_2}=\frac{0.2^2}{0.126^2}=2.520[/tex]
Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have [tex]n_1 -1 =10-1=9[/tex] and for the denominator we have [tex]n_2 -1 =20-1=19[/tex] and the F statistic have 9 degrees of freedom for the numerator and 19 for the denominator. And the P value is given by:
Now we can calculate the p value with this probability:
[tex]p_v =P(F_{9,19}>2.520)=0.043[/tex]
Using a significance level of 5% we see that the p value is lower than this value and we have enough evidence to reject the null hypothesis and we can conclude that the variation for the new process is lower than the new one.
Step-by-step explanation:
Information given
[tex]n_1 = 10 [/tex] represent the sampe size old
[tex]n_2 =20[/tex] represent the sample size new
[tex]s_1 = 0.2[/tex] represent the sample deviation for old
[tex]s_2 = 0.126[/tex] represent the sample deviation for new
The statistic is given by:
[tex]F=\frac{s^2_1}{s^2_2}[/tex]
Hypothesis to test
We want to test if the new process is less variable than the old, so the system of hypothesis are:
H0: [tex] \sigma^2_1 \leq \sigma^2_2[/tex]
H1: [tex] \sigma^2_1 >\sigma^2_2[/tex]
The statistic is given by:
[tex]F=\frac{s^2_1}{s^2_2}=\frac{0.2^2}{0.126^2}=2.520[/tex]
Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have [tex]n_1 -1 =10-1=9[/tex] and for the denominator we have [tex]n_2 -1 =20-1=19[/tex] and the F statistic have 9 degrees of freedom for the numerator and 19 for the denominator. And the P value is given by:
Now we can calculate the p value with this probability:
[tex]p_v =P(F_{9,19}>2.520)=0.043[/tex]
Using a significance level of 5% we see that the p value is lower than this value and we have enough evidence to reject the null hypothesis and we can conclude that the variation for the new process is lower than the new one.
To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assigned to the three treatments. You are given the results below. Treatment Observation A 20 30 25 33 B 22 26 20 28 C 40 30 28 22 The test statistic to test the null hypothesis equals _____.
Answer:
The test statistic to test the null hypothesis equals 1.059
Step-by-step explanation:
From the given information; we have:
Treatment Observations
A 20 30 25 33
B 22 26 20 28
C 40 30 28 22
The objective is to find the test statistic to test the null hypothesis; in order to do that;we must first run through a series of some activities.
Let first compute the sum of the square;
Total sum of squares (TSS) = Treatment sum of squares [tex](T_r SS)[/tex] + Error sum of squares (ESS)
where:
(TSS) = [tex]\sum \limits ^v_{i=1} \sum \limits ^{n_i}_{j-1}(yij- \overline {y}oo)^2[/tex] with (n-1) df
[tex](T_r SS)[/tex] [tex]= \sum \limits ^v_{i=1} n_i( \overline yio- \overline {y}oo)^2[/tex] with (v-1) df
[tex](ESS) = \sum \limits ^v_{i=1} \sum \limits ^{n_i}_{j-1}(yij- \overline {y}io)^2[/tex] with (n-v) df
where;
v= 3
[tex]n_i=[/tex]4
i = 1,2,3
n =12
[tex]y_{ij}[/tex] is the [tex]j^{th[/tex] observation for the [tex]i^{th[/tex] treatment
[tex]\overline{y}io[/tex] is the mean of the [tex]i^{th[/tex] treatment i = 1,2,3 ; j = 1,2,3,4
[tex]\overline y oo[/tex] is the overall mean
From the given data
[tex]\overline y oo = \dfrac{1}{12} \sum \limits ^3_{i=1} \sum \limits ^{4}_{j=1}(yij)^2= 27[/tex]
[tex]TSS = \dfrac{1}{12} \sum \limits ^3_{i=1} \sum \limits ^{4}_{j=1}(yij- 27)^2 = 378[/tex]
[tex]T_r SS= \sum \limits^3_{i=1}4 (\overline y io - \overline yoo)^2[/tex]
[tex]=4(27-27)^2+4(24-27)^2+4(30-27)^2 = 72[/tex]
Total sum of squares (TSS) = Treatment sum of squares [tex](T_r SS)[/tex] + Error sum of squares (ESS)
(TSS) = 378 - 72
(TSS) = 306
Now; to the mean square between treatments (MSTR); we use the formula:
MSTR = TrSS/df(TrSS)
MSTR = 72/(3 - 1)
MSTR = 72/2
MSTR = 36
The mean square within treatments (MSE) is:
MSE = ESS/df(ESS)
MSE = 306/(12-3)
MSE = 306/(9)
MSE = 34
The test statistic to test the null hypothesis is :
[tex]T = \dfrac{ \dfrac{TrSS}{\sigma^2}/(v-1) }{ \dfrac{ESS}{\sigma^2}/(n-v) } = \dfrac{MSTR}{MSE} \ \ \ \approx \ \ T(\overline {v-1}, \overline {n-v})[/tex]
[tex]T = \dfrac{36}{34}[/tex]
T = 1.059
Whats the theorum called for working out the missing side of a triangle?
Answer:
The Pythagorean Theorum
Step-by-step explanation:
you can literally search it and see that it is right!
Solve the following equation. x + 6 = x + x
Answer:
x = 6
Step-by-step explanation:
x + 6 = x + x
Combine like terms
x+6 =2x
Subtract x from each side
x+6-x = 2x-x
6 = x
I don’t know this one
Answer:
C
Step-by-step explanation:
2/3x - 5>3
Add 5 to each side
2/3x - 5+5>3+5
2/3x > 8
Multiply each side by 3/2
3/2 *2/3x > 8*3/2
x > 12
There is an open circle at 12 and the lines goes to the right
Given the two parallel lines determine the value of x
Answer:
D. 150°
Step-by-step explanation:
x= 150°
Choice D
600000000*100000000000000000000000000000000000000000000
Answer:
6e+52
Step-by-step explanation:
cAlCuLaToR
Answer:
6e+52
Step-by-step explanation:
multiply
Solve the equation 3 Z + 5 = 35
Answer:
z=10 i hope this will help you
Step-by-step explanation:
3z+5=35
3z=35-5
3z=30
z=10
Answer:
Z = 10
Step-by-step explanation:
3Z+5=35
Subtract 5 from both sides
3Z=30
Divide both sides by 3
Z=10
Composition of the function is commuatative
Answer:
The functions g and f are said to commute with each other if g ∘ f = f ∘ g. Commutativity is a special property, attained only by particular functions, and often in special circumstances. For example, |x| + 3 = |x + 3| only when x ≥ 0. ... The composition of one-to-one functions is always one-to-one.
\
There are 15 marbles in a bag; 10 are blue, 4 are red and 1 is green. Marbles are drawn and NOT replaced 8 times, with the number of red marbles being recorded. What is the probability of getting exactly 3 red marbles? (Write as a percentage, correct to two decimals. eg: 12.34%)
Answer: There is a 0.88% chance of pulling three red marbles in a row.
Step-by-step explanation:
First pull = 4/15 (26.67%)
second pull = 3/14 (21.43%)
Third pull = 2/13 (15.38%)
You need to multiply these three fractions to get the probability of pulling three reds in a row, doing that will get you 4/455 or 0.88%
If f(x) = (-x)^3, what is f(-2)?
-6
-8
8
6
Answer:
The answer is 8
Step-by-step explanation:
Plug -2 in for x. The double-negative inside the parenthesis makes it positive, then do the exponent.
Answer:
-(-2)^3 = 2^3 = 8
Answer is C
Step-by-step explanation:
So we plug in the numbers. We have -2 as x. (-(-2)^3 would be our thing. Thats because our x is the negative so the negative of -2 is 2.
2^3 = 8
therefore its 8
Problem PageQuestion A Web music store offers two versions of a popular song. The size of the standard version is 2.6 megabytes (MB). The size of the high-quality version is 4.2 MB. Yesterday, the high-quality version was downloaded four times as often as the standard version. The total size downloaded for the two versions was 4074 MB. How many downloads of the standard version were there?
Answer:
There were 210 downloads of the standard version.
Step-by-step explanation:
This question can be solved using a system of equations.
I am going to say that:
x is the number of downloads of the standard version.
y is the number of downloads of the high-quality version.
The size of the standard version is 2.6 megabytes (MB). The size of the high-quality version is 4.2 MB. The total size downloaded for the two versions was 4074 MB.
This means that:
[tex]2.6x + 4.2y = 4074[/tex]
Yesterday, the high-quality version was downloaded four times as often as the standard version.
This means that [tex]y = 4x[/tex]
How many downloads of the standard version were there?
This is x.
[tex]2.6x + 4.2y = 4074[/tex]
Since [tex]y = 4x[/tex]
[tex]2.6x + 4.2*4x = 4074[/tex]
[tex]19.4x = 4074[/tex]
[tex]x = \frac{4074}{19.4}[/tex]
[tex]x = 210[/tex]
There were 210 downloads of the standard version.
According to 2013 report from Population Reference Bureau, the mean travel time to work of workers ages 16 and older who did not work at home was 30.7 minutes for NJ State with a standard deviation of 23 minutes. Assume the population is normally distributed.
Required:
a. If a worker is selected at random, what is the probability that his travel time to work is less than 30 minutes?
b. Specify the mean and the standard deviation of the sampling distribution of the sample means, for samples of size 36.
c. What is the probability that in a random sample of 36 NJ workers commuting to work, the mean travel time to work is above 35 minutes?
Answer:
a) 48.80% probability that his travel time to work is less than 30 minutes
b) The mean is 30.7 minutes and the standard deviation is of 3.83 minutes.
c) 13.13% probability that in a random sample of 36 NJ workers commuting to work, the mean travel time to work is above 35 minutes
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question, we have that:
[tex]\mu = 30.7, \sigma = 23[/tex]
a. If a worker is selected at random, what is the probability that his travel time to work is less than 30 minutes?
This is the pvlaue of Z when X = 30. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{30 - 30.7}{23}[/tex]
[tex]Z = -0.03[/tex]
[tex]Z = -0.03[/tex] has a pvalue of 0.4880.
48.80% probability that his travel time to work is less than 30 minutes
b. Specify the mean and the standard deviation of the sampling distribution of the sample means, for samples of size 36.
[tex]n = 36[/tex]
Applying the Central Limit Theorem, the mean is 30.7 minutes and the standard deviation is [tex]s = \frac{23}{\sqrt{36}} = 3.83[/tex]
c. What is the probability that in a random sample of 36 NJ workers commuting to work, the mean travel time to work is above 35 minutes?
This is 1 subtracted by the pvalue of Z when X = 35. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{35 - 30.7}{3.83}[/tex]
[tex]Z = 1.12[/tex]
[tex]Z = 1.12[/tex] has a pvalue of 0.8687
1 - 0.8687 = 0.1313
13.13% probability that in a random sample of 36 NJ workers commuting to work, the mean travel time to work is above 35 minutes
The following histogram shows the exam scores for a Prealgebra class. Use this histogram to answer the questions.Prealgebra Exam ScoresScores 70.5, 75.5, 80.5, 85.5, 90.5, 95.5, 100.5Frequency 0, 4, 8, 12, 16, 20, 24Step 1 of 5:Find the number of the class containing the largest number of exam scores (1, 2, 3, 4, 5, or 6).Step 2 of 5:Find the upper class limit of the third class.Step 3 of 5:Find the class width for this histogram.Step 4 of 5:Find the number of students that took this exam.Step 5 of 5:Find the percentage of students that scored higher than 95.595.5. Round your answer to the nearest percent.
Answer:
The number of the class containing the largest score can be found in frequency 24 and the class is 98 - 103
For the third class 78 - 83 ; the upper limit = 83
The class width for this histogram 5
The number of students that took the exam simply refers to the frequency is 84
The percentage of students that scored higher than 95.5 is 53%
Step-by-step explanation:
The objective of this question is to use the following histogram that shows the exam scores for a Pre-algebra class to answer the question given:
NOW;
The table given in the question can be illustrated as follows:
S/N Class Score Frequency
1 68 - 73 70.5 0
2 73 - 78 75.5 4
3 78 - 83 80.5 8
4 83 - 88 85.5 12
5 88 - 93 90.5 16
6 93 - 98 95.5 20
7 98 - 103 100.5 24
TOTAL: 84
a) The number of the class containing the largest score can be found in frequency 24 and the class is 98 - 103
b) For the third class 78 - 83 ; the upper limit = 83 ( since the upper limit is derived by addition of 5 to the last number showing in the highest value specified by the number in the class interval which is 78 ( i.e 78 + 5 = 83))
c) The class width for this histogram 5 ; since it is the difference between the upper and lower boundaries limit of the given class.
So , from above the difference in any of the class will definitely result into 5
d) The number of students that took the exam simply refers to the frequency ; which is (0+4+8+12+16+20+24) = 84
e) Lastly; the percentage of students that scored higher than 95.5 is ;
⇒[tex]\dfrac{20+24}{84} *100[/tex]
= 0.5238095 × 100
= 52.83
To the nearest percentage ;the percentage of students that scored higher than 95.5 is 53%
Answer:
1. 98-103 (6th class)
2. 88
3. 5
4. 84
5. 52%
Step-by-step explanation:
Find attached the frequency table.
The class of exam scores falls between (1, 2, 3, 4, 5, or 6).
The exam score ranged from 68-103
1) The largest number of exam scores = 24
The largest number of exam scores is in the 6th class = 98 -103
Step 2 of 5:
The upper class limit is the higher number in an interval. Third class interval is 83-88
The upper class limit of the third class 88.
Step 3 of 5:
Class width = upper class limit - lower class limit
We can use any of the class interval to find this as the answer will be the same. Using the interval between 73-78
Class width = 78 - 73
Class width for the histogram = 5
Step 4 of 5:
The total of students that took the test = sum of all the frequency
= 0+4+8+12+16+20+24 = 84
The total of students that took the test = 84
Step 5 of 5:Find the percentage of students that scored higher than 95.5
Number of student that scored higher than 95.5 = 20 + 24 = 44
Percentage of students that scored higher than 95.5 = [(Number of student that scored higher than 95.5)/(total number of students that took the test)] × 100
= (44/84) × 100 = 0.5238 × 100 = 52.38%
Percentage of students that scored higher than 95.5 = 52% (nearest percent)
Use technology to approximate the solution(s) to the system of equations to the nearest tenth of a unit. Select all that apply.
{f(x) = 2(3)^x
{g(x) = 10log(x+3)
(-1.9, 15.9)
(-2, 0.2)
(1.9, -15.9)
(2, -0.2) (1.9, 15.9)
Answer:
closest choice: (-2, 0.2)
Step-by-step explanation:
The attached image from a graphing calculator shows the solutions (to the nearest tenth) to be ...
(-1.9, 0.2)
(1.0, 6.0)
The closest of the offered choices is (-2, 0.2). None are actually correct.
A rectangle has an area of 96 cm2 The length of the rectangle is 4 cm longer than the width. Work out the length and width of the rectangle.
A rectangular fish tank is 50 cm long, 40 cm wide, and 20 cm high. a) How many cubic centimeters of water will the tank hold? b) How many milliliters of water will the tank hold? c) How many liters of water will the tank hold?
Answer:
40 litres
Step-by-step explanation:
V = l x w x h
50 x 40 x 20 = 40000
40000 cm^3
1cm^3 = 1ml
40000 cm^3/ 1cm^3 = 40000ml
40000 x 10^-3 = 40 litres
Mathematics
ose the correct answer:
. What number should be added to (-5/16) to get ( 7/24)?
Answer:
0.6042 or 29/48
Step-by-step explanation:
-5/16 = -0.3125
7/24 = 0.2917
0.2917 - -0.3125 = 0.6042
0.6042 ≅ 29/48
Answer:
29/48
Step-by-step explanation:
-5/16 + x= 7/24
x= 7/24-(-5/16)
x=7/24+5/16
x= 2*7/2*24+ 3*5/3*16
x=29/48
of 5 points)
2. The two figures are similar. Write the similarity statement. Justify you
37.5
(Score for Question 2:
45
Y
40
30
Z
Answer:
Answer:
f\left(x\right)=x^3-x
Step-by-step explanation:
The figure shows three lines that intersect at point N.
3 lines intersect. Lines G K and J M intersect at point N to form a right angle. Line H L intersects the other lines at point N. Angle H N G is 48 degrees.
Angle GNH is congruent to angle KNL. Angle MNL is complementary to angle KNL. What is the measure of angle MNL?
42°
48°
132°
138°
Answer
The figure shows three lines that intersect at point N.
3 lines intersect. Lines G K and J M intersect at point N to form a right angle. Line H L intersects the other lines at point N. Angle H N G is 48 degrees.
Angle GNH is congruent to angle KNL. Angle MNL is complementary to angle KNL. What is the measure of angle MNL?
42°
48°
132°
138°
Step-by-step explanation:
42
The required measure of angle MNL is 42°. Option A is correct.
3 lines intersect lines G K and J M intersect at point N to form a right angle. Line H L intersects the other lines at point N. Angle H N G is 48 degrees. Angle GNH is congruent to angle KNL. Angle MNL is complementary to angle KNL. What is the measure of angle MNL is to determine
The angle can be defined as the one line inclined over another line.
unit of measure of an angle is degree and radians.
Angle GNH is congruent to angle KNL.
∠KNL = 48°
Angle MNL is complementary to angle KNL
Since angle MNK = 90°
∠MNL + ∠KNL = 90°
∠MNL = 90-48
∠MNL = 42°
Thus, the required measure of angle MNL is 42°. Option A is correct.
Learn more about Angles here:
https://brainly.com/question/13954458
#SPJ6
The point A (-7,5) is reflected over the line x = -5, and then is reflected over the line x= 2. What are the coordinates of
A?
o (7, 19)
O (10,5)
(7,5)
(10, 19)
Answer:
(7, 5) is the final reflection of the point.
Step-by-step explanation:
We are given point A(-7, 5) which is first reflected over the line [tex]x= -5[/tex].
The minimum distance of the point A(-7, 5) from the line [tex]x= -5[/tex] is 2 units across the horizontal path (No change in y coordinate).
Point A lies 2 units on the left side of the line [tex]x= -5[/tex].
So, its reflection will be 2 units on the right side of [tex]x= -5[/tex].
Let its reflection be A' which has coordinates as (-5+2,5) i.e. (-3, 5).
Now A'(-3, 5) is reflected on the line [tex]x=2[/tex].
The minimum distance of the point A'(-3, 5) from the line [tex]x=2[/tex] is 5 units across the horizontal path (No change in y coordinate).
Point A' lies 5 units on the left side of the line [tex]x=2[/tex].
So, its reflection will be 5 units on the right side of [tex]x=2[/tex].
Let its reflection be A'' which has coordinates as (2+5, 5) i.e (7, 5) is the final reflection of the point..
Please find attached image.
(7, 5) is the final reflection of the point.
According to the Bureau of Labor Statistics, 7.1% of the labor force in Wenatchee, Washington was unemployed in February 2019. A random sample of 100 employable adults in Wenatchee, Washington was selected. Using the normal approximation to the binomial distribution, what is the probability that 6 or more people from this sample are unemployed
Answer:
73.24% probability that 6 or more people from this sample are unemployed
Step-by-step explanation:
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]n = 100, p = 0.071[/tex]
So
[tex]\mu = E(X) = np = 10*0.071 = 7.1[/tex]
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.071*0.929} = 2.5682[/tex]
What is the probability that 6 or more people from this sample are unemployed
Using continuity correction, this is [tex]P(X \geq 6 - 0.5) = P(X \geq 5.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 5.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5.5 - 7.1}{2.5682}[/tex]
[tex]Z = -0.62[/tex]
[tex]Z = -0.62[/tex] has a pvalue of 0.2676
1 - 0.2676 = 0.7324
73.24% probability that 6 or more people from this sample are unemployed
01:30:4
Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26. What is the
solution set of this problem?
Answer:
x<_ 21
Step-by-step explanation:
5(x+27)>_ =6(x+26)
5x +135 >_ 6x +156
5x >_6x +21
-x>_21
x<_21
Determine whether the following sequence converges or diverges and describe whether it does do so monotonically or by oscillation. Give the limit when the sequence converges.
{(-1.00000005)^n}
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. The sequence diverges by oscillation.
b. The sequence converges monotonically. It converges to:________
c. The sequence converges by oscillation. It converges to:________
d. The sequence diverges monotonic ally.
Answer:
a
Step-by-step explanation:
(-1.00000005)^n
as n becomes very large, the function increases in both positive and negative direction.
If n=1, -1.00000005
if n=2, 1.0000001
if n= 3, -1.00000015
if n=20, 1.000001
if n=21, -1.00000105
The random variable x represents the number of computers that families have along with the corresponding probabilities. Find the mean and standard deviation for the random variable x.
Answer:
The correct option is (d).
Step-by-step explanation:
The complete question is:
The random variable x represents the number of computers that families have along with the corresponding probabilities. Use the probability distribution table below to find the mean and standard deviation for the random variable x.
x : 0 1 2 3 4
p (x) : 0.49 0.05 0.32 0.07 0.07
(a) The mean is 1.39 The standard deviation is 0.80
(b) The mean is 1.39 The standard deviation is 0.64
(c)The mean is 1.18 The standard deviation is 0.64
(d) The mean is 1.18 The standard deviation is 1.30
Solution:
The formula to compute the mean is:
[tex]\text{Mean}=\sum x\cdot p(x)[/tex]
Compute the mean as follows:
[tex]\text{Mean}=\sum x\cdot p(x)[/tex]
[tex]=(0\times 0.49)+(1\times 0.05)+(2\times 0.32)+(3\times 0.07)+(4\times 0.07)\\\\=0+0.05+0.64+0.21+0.28\\\\=1.18[/tex]
The mean of the random variable x is 1.18.
The formula to compute variance is:
[tex]\text{Variance}=E(X^{2})-[E(X)]^{2}[/tex]
Compute the value of E (X²) as follows:
[tex]E(X^{2})=\sum x^{2}\cdot p(x)[/tex]
[tex]=(0^{2}\times 0.49)+(1^{2}\times 0.05)+(2^{2}\times 0.32)+(3^{2}\times 0.07)+(4^{2}\times 0.07)\\\\=0+0.05+1.28+0.63+1.12\\\\=3.08[/tex]
Compute the variance as follows:
[tex]\text{Variance}=E(X^{2})-[E(X)]^{2}[/tex]
[tex]=3.08-(1.18)^{2}\\\\=1.6876[/tex]
Then the standard deviation is:
[tex]\text{Standard deviation}=\sqrt{\text{Variance}}[/tex]
[tex]=\sqrt{1.6876}\\\\=1.2990766\\\\\approx 1.30[/tex]
Thus, the mean and standard deviation for the random variable x are 1.18 and 1.30 respectively.
The correct option is (d).
There are 4 blue tiles, 12 red tiles, and 6 green tiles in a bag. Which model represents the probability, P, that Luke will pick a red tile from the bag?
Answer:
The Probability that will pick a red tile from the bag
[tex]P(E) = \frac{6}{11}[/tex] = 0.545
Step-by-step explanation:
Explanation:-
Given data 4 blue tiles, 12 red tiles, and 6 green tiles in a bag
Total = 4 B + 12 R + 6 G = 22 tiles
Total number of exhaustive cases
n (S) = [tex]22 C_{1} = 22 ways[/tex]
The Probability that will pick a red tile from the bag
[tex]P(E) = \frac{n(E)}{n(S)} = \frac{12 C_{1} }{22 C_{1} } = \frac{12}{22}[/tex]
[tex]P(E) = \frac{6}{11}[/tex]
P(E) = 0.545
Final answer:-
The Probability that will pick a red tile from the bag = 0.545
What’s the correct answer for this question?
Answer:
B:
Step-by-step explanation:
According to theorem, "the angle in a semi-circle is a right angle" So,
<O = 90°
<M = 54
<K = 180-90-54
<OKM = 36°
The cost of a circular table is directly proportional to the square of the radius. A circular table with a radius of 50cm costs £60. What is the cost of a circular table with a radius of 75cm? Show all your working
Answer:
£135 is the correct answer.
Step-by-step explanation:
Let C be the cost of table.
And let R be the radius of table.
Cost of table is directly proportional to square of radius.
As per question statement:
[tex]C\propto R^{2}[/tex] or
[tex]C=a\times R^2 ....... (1)[/tex]
where [tex]a[/tex] is the constant to remove the [tex]\propto sign[/tex].
It is given that
[tex]C_1 =[/tex] £60 and [tex]R_1 = 50\ cm[/tex]
[tex]C_2 = ?[/tex] when [tex]R_2= 75\ cm[/tex]
Putting the values of [tex]C_1[/tex] and [tex]R_1[/tex] in equation (1):
[tex]60=a \times 50^2 ....... (2)[/tex]
Putting the values of [tex]C_2[/tex] and [tex]R_2[/tex] in equation (1):
[tex]C_2=a \times 75^2 ....... (3)[/tex]
Dividing equation (2) by (3):
[tex]\dfrac{60}{C_2}= \dfrac{a \times 50^2}{a \times 75^2}\\\Rightarrow \dfrac{60}{C_2}= \dfrac{50^2}{75^2}\\\Rightarrow \dfrac{60}{C_2}= \dfrac{2^2}{3^2}\\\Rightarrow \dfrac{60}{C_2}= \dfrac{4}{9}\\\Rightarrow C_2 = 15 \times 9 \\\Rightarrow C_2 = 135[/tex]
So, £135 is the correct answer.
HELP PLEASE!!
NEED ANSWER ASAP!!!
A farmer in China discovers a mammal
hide that contains 54% of its original
Find age of the mammal hide to the nearest year.
amount of C-14
N=N0e^-kt
N = Noe
No = inital amount of C-14 (at time t = 0)
N = amount of C-14 at time t
k = 0.0001
t = time, in years
Answer:
6163.2 years
Step-by-step explanation:
A_t=A_0e^{-kt}
Where
A_t=Amount of C 14 after “t” year
A_0= Initial Amount
t= No. of years
k=constant
In our problem we are given that A_t is 54% that is if A_0=1 , A_t=0.54
Also , k=0.0001
We have to find t=?
Let us substitute these values in the formula
0.54=1* e^{-0.0001t}
Taking log on both sides to the base 10 we get
log 0.54=log e^{-0.0001t}
-0.267606 = -0.0001t*log e
-0.267606 = -0.0001t*0.4342
t=\frac{-0.267606}{-0.0001*0.4342}
t=6163.20
t=6163.20 years
PLEASE MARK BRAINLY
HELP PLEASE SIMPLIFY !!!
Answer:
[tex]=x^{\frac{5}{6}}+2x^{\frac{7}{3}}[/tex]
Step-by-step explanation:
[tex]x^{\frac{1}{3}}\left(x^{\frac{1}{2}}+2x^2\right)\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac\\a=x^{\frac{1}{3}},\:b=x^{\frac{1}{2}},\:c=2x^2\\=x^{\frac{1}{3}}x^{\frac{1}{2}}+x^{\frac{1}{3}}\cdot \:2x^2\\=x^{\frac{1}{3}}x^{\frac{1}{2}}+2x^2x^{\frac{1}{3}}\\\mathrm{Simplify}\:x^{\frac{1}{3}}x^{\frac{1}{2}}+2x^2x^{\frac{1}{3}}:\quad x^{\frac{5}{6}}+2x^{\frac{7}{3}}\\x^{\frac{1}{3}}x^{\frac{1}{2}}+2x^2x^{\frac{1}{3}}\\x^{\frac{1}{3}}x^{\frac{1}{2}}=x^{\frac{5}{6}}[/tex]
[tex]x^{\frac{1}{3}}x^{\frac{1}{2}}\\\mathrm{Apply\:exponent\:rule}:\quad \:a^b\cdot \:a^c=a^{b+c}\\x^{\frac{1}{3}}x^{\frac{1}{2}}=\:x^{\frac{1}{3}+\frac{1}{2}}\\=x^{\frac{1}{3}+\frac{1}{2}}\\\mathrm{Join}\:\frac{1}{3}+\frac{1}{2}:\quad \frac{5}{6}\\\frac{1}{3}+\frac{1}{2}\\\mathrm{Least\:Common\:Multiplier\:of\:}3,\:2:\quad 6\\Adjust\:Fractions\:based\:on\:the\:LCM\\=\frac{2}{6}+\frac{3}{6}[/tex]
[tex]\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}\\=\frac{2+3}{6}\\\mathrm{Add\:the\:numbers:}\:2+3=5\\=\frac{5}{6}\\=x^{\frac{5}{6}}\\2x^2x^{\frac{1}{3}}=2x^{\frac{7}{3}}\\=x^{\frac{5}{6}}+2x^{\frac{7}{3}}[/tex]
In a group of 50 patrons, 14 patrons like lattes and espressos, 11 patrons like
espressos and cappuccinos, 7 patrons like lattes and cappuccinos, and 3
patrons like all 3 coffee drinks. Altogether, 22 patrons like lattes, 30 patrons
like espressos, and 23 patrons like cappuccinos. How many patrons don't like
any of these coffee drinks?
Answer:the answer would be 4. Hope this helps.
Step-by-step explanation:
Using the formula of union of three events, the number of patrons who didn't like any of given coffee drinks = 4.
What is union of three events?Union of three events : P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C).
n (latte ∩ espressos) = 14
n (espressos ∩ cappuccinos) = 11
n (lattes ∩ cappuccinos) = 7
n (latte ∩ espressos ∩ cappuccinos) = 3
n (lattes) = 22
n (espressos) = 30
n (cappuccinos) = 23
n(latte ∪ espressos ∪ cappuccinos) =
= n (lattes) + n (espressos) + n (cappuccinos) - n (latte ∩ espressos) - n (espressos ∩ cappuccinos) - n (lattes ∩ cappuccinos) + n (latte ∩ espressos ∩ cappuccinos)
= 22 + 30 + 23 - 14 - 11 - 7 + 3
= 46
n (universe) = 50
Number of patrons who didn't like any of these drinks =
= n (universe) - n (latte ∪ espressos ∪ cappuccinos) = 50 - 46 = 4
Learn more about union of three events here
https://brainly.com/question/14614116
#SPJ3