Solve the systems of equations: 3x+2y=4
-2x+2y=24
Please show work for how you solved for x and y

The probability that the mean number of hours spent studying is less than 70 hours is; 0.023

We are given that:

Population mean; μ = 75

Population Standard deviation; σ = 25

Sample size; n = 100

The formula for the z-score is;

z = (x' - μ)/(σ/√n)

We want to find P(x < 70). Thus;

z = (70 - 75)/(25/√100)

z = -5/2.5

z = -2

From online p-value from z-score calculator, we have;

P(Z < -2) = 0.023

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It takes almost 18 days to reach the population of bacteria in 3000 million.

What is the growth of bacteria?

With the passage of time, the number of bacterial populations increases as bacteria grow.

Why does growth of bacteria count is necessary?

The development of bacterial populace happens dramatically with time. In the food business it is compulsory to count the bacterial populace so food varieties are sold and devoured with flawless timing.

According to the given question:

Given, population of bacteria in millions after t days of packing= 3000

The number of bacteria is given by the equation, f(t) = 500 e∧0.1t

where t is the time in day.

we need to calculate the time taken to grow 3000 million bacteria

f(t) =3000

from the above equation, 3000 = 500e∧0.1t

                                     3000/500 = e∧ 0.1t

                                    6 = e∧0.1t

                                    ㏑6 = ㏑(e∧0.1t)     [take ln on both side]

                                     1.7918 = 0.1t

                                  time, t= 17.92 days or 18 days

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If f(x,y,z) is a vector field rather than a scalar function. A vector-valued function is one that has a vector value and a calculable curl. Therefore, if we calculate this value, it will once more be a vector-valued function. The right response in this case is option B.

The following considerations must be made in order to answer this question: It is specified that the divergence of a vector field is a scalar-valued function.

A vector field's specified and actual curl is a vector field. A scalar field's divergence/Curl is not specified. A scalar field's gradient is a vector field that has a definite definition. A vector field's gradient is not known.

Since F is a vector-valued function, we may discover its divergence, which will be a scalar-valued function, and since the gradient of the scalar-valued function will be a vector-valued function, this calculation is attempting to compute the gradient of the divergence of the vector field F.

Complete question:

Let f(x,y,z) be a scalar-valued function, and F(x,y,z) be a vector field. Is the following defined? If it is defined, is it a scalar-valued function or a vector field? (You are only allowed one attempt for each part of this problem.)

(a) V (Vn Defined- it is a scalar-valued function O Defined - it is a vector field O Not defined

(b) VV F) O Defined - it is a scalar-valued function Defined - it is a vector field O Not defined

(c) V (V F) Defined it is a scalar-valued function Defined - it is a vector field Not defined

(d) V (V x F) Defined it is a scalar-valued function Defined it is a vector field Not defined

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12090 people are there after 30 days when the growth rate of the population of a culture of bacteria, p(t).

Given that,

The growth rate of the population of a culture of bacteria, p(t), where t is the number of days, is 0.2 and is proportionate to the population's growth. Initially, there are 30 people.

We have to find how many people are there after 30 days. (Stop rounding your response.) and when will there be a population doubling.

We know that,

dp//dt directly proportional p

dp/dt=kp

Here,

k=0.2

dp/p=0.2dt

Integrating on both sides.

[tex]\int {1/p} \, dp =\int {0.2} \, dt[/tex]

logp=0.2t+c

p=c[tex]e^{0.2t}[/tex]

Here,

t=30

c=30

So,

p=30[tex]e^{0.2(30)}[/tex]

p=30×403

p=12090

Therefore, 12090 people are there after 30 days when the growth rate of the population of a culture of bacteria, p(t).

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99 many edges must be included in any tree with 100 vertices.

This is because, in a tree, there must be one fewer edge than vertices. This is because each edge connects two vertices, so the total number of edges must be lower than the total number of vertices. Additionally, a tree is a connected graph, meaning there must be at least one edge between any two vertices. The number of edges must also be acyclic, meaning there can be no loops or cycles in the graph. Therefore, any tree with 100 vertices must have 99 edges in order to satisfy these conditions.

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If it is important to detect a mean difference in score of one point with the probability of at least 0.90 , then the number of pairs that should be used is  10 .

It is given that ,

the probability is at least 0.90 ,

So , in the paired t test ,

testing mean paired difference is = 0

alpha = 0.05 , the assumed standard deviation of paired difference = 0.441

So the output is

Difference = 1 , Size = 5 power(probability) = 0.90 ,

So , the actual probability is = 0.95190

From the output above , the required sample size is n = 5 .

We observe that , under the given conditions the sample size is = 5 .

but the researcher  considered 10 pairs ,

So , the sample size 10 is used for this study .

Therefore , 10 pairs should be used .

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The given question is incomplete ,the complete question is

It is important to detect a mean difference in score of one point with a probability of at least 0.90. how many pairs should have been used ?

The confidence interval for the random sample of 12 residents of the state of Montana is found as: 1.764 ≤ μ ≤ 2.636.

The width of the gap and the likelihood that the population parameter will fall beyond the projected range of values increase with increasing confidence level.

The stated data;

Degree of freedom

Df = n - 1

Df = 12 - 1

Df = 11

The critical value of t.

t critical = t(α/2,Df)

             = t(0.05, 11)

Using the t distribution.

t critical = ± 1.796

The confidence interval:

μ = x ± t.s / √n

 = 2.2 ± (1.796).(0.84)/√12

= 2.2 + 0.4355

1.764 ≤ μ ≤ 2.636

Thus, the confidence interval for the random sample of 12 residents of the state of Montana is found as: 1.764 ≤ μ ≤ 2.636.

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By finding the combinations C(15, 3), we wills see that there are 455 different sets of prizes.

Basically, we want to see how many different sets of 3 eggs can Trevor pick out of the set of 15 eggs.

So we want to find the combinations, remember that for a set of N elements, the number of different subsets of K elements is given by:

C(N. K) = N!/(K!*(N - K)!)

Here we have:

N = 15

K = 3

Then:

C(15, 3) = 15!/(3!*12!) = 15*14*13/3*2 = 455

There are 455 different sets of prizes that Trevor could pick.

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There are 455 different sets of prizes.

True, Building more movie theaters will cause the homicide rate to rise.There is a significant linear correlation.

What does a statistical correlation mean?

A statistical metric known as correlation, which is given as a number, describes the strength and direction of a relationship between two or more variables.

                Although there may be a correlation between two variables, this does not necessarily imply that the change in one variable is what led to the change in the values of the other variable.

Correlation does not always indicate cause. Many factors, such as multi colinearity between movie theaters and the number of killings, which both tend to rise with population, can contribute to a strong connection. In general, there are numerous factors that could account for the parallels between the two.

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Answer:

Lisa's original weight was 172.9 pounds.

Step-by-step explanation:

Let w be Lisa's original weight and l be the weight she lost. Since Lisa lost 27% of her original weight, we have:

0.27w = l

Substituting the known value of l, we get:

0.27w = 46.71

w = 46.71 / 0.27

w = 172.9

Thus, Lisa's original weight was 172.9 pounds.

The average wait time to see an E.R. doctor is significantly less than 150 minutes, and we can recommend to the winery that the wait time is actually less than what is reported.

To test the hypothesis that the average wait time to see an E.R. doctor is less than 150 minutes, we can follow these steps:

State the null hypothesis: The null hypothesis is the assumption that there is no difference between the observed result and what we expect to see. In this case, the null hypothesis is that the average wait time to see an E.R. doctor is 150 minutes.

State the alternative hypothesis: The alternative hypothesis is the opposite of the null hypothesis. In this case, the alternative hypothesis is that the average wait time to see an E.R. doctor is less than 150 minutes.

Calculate the test statistic: The test statistic is a measure of the difference between the observed result and the expected result. In this case, the test statistic is the difference between the observed average wait time of 148 minutes and the expected average wait time of 150 minutes, divided by the standard deviation of 5 minutes.

Determine the critical value: The critical value is the value that determines whether the test statistic is significant or not. To determine the critical value, we need to determine the p-value of the test statistic using a significance level of 0.05.

Compare the test statistic to the critical value: If the test statistic is greater than the critical value, then the null hypothesis can be rejected. If the test statistic is less than the critical value, then the null hypothesis cannot be rejected.

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Based on the following graph, The hippocampus in the brain is the most likely site of the lesion.

The brain is a sophisticated organ that manages every bodily function as well as thought, memories, emotions, touch, motor function, vision, respiration, temperature, and hunger. Its central nervous system or CNS, is made up of the spinal cord that emerges from the brain.

The brain regulates many bodily functions, including the function of numerous organs as well as thinking, memory, language, arm and leg motions. By controlling heart and breathing rates, it also affects how people react to stressful situations (such as completing an exam, losing a job, having a baby, being ill, etc.).

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Answer:

  120

Step-by-step explanation:

You want to know the measure of the third arc of a circle where one of the arcs is labeled 114°, and a second arc is intercepted by a 63° inscribed angle.

The measure of an inscribed angle (63°) is half the measure of the arc it intercepts:

  (1/2)(arc) = 63°

  arc = 2·63° = 126° . . . . . arc opposite the 63° angle

The total of arcs of a circle total 360°.

  114° +126° +x° = 360°

  x° = 120° . . . . . . . . . subtract 240°

The value of x is 120.

1) On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar,

a) 90% CI for the true average yield point of the modified bar is 8459 +/- 23.5

b) 98% CI for the true average yield point of the modified bar is 8459 +/- 23.49

2) An article reported that for a sample of kitchens with gas cooking appliances monitored during a one-week period,

a) 95% (two-sided) confidence interval for true average CO2 level in the population is 654.16 +/- 185.0632

b) Sample size is 12.

The confidence interval for the population mean is constructed about the sample mean i.e. sample mean lies at the center of the interval.

1) The yield point of a particular type of mild steel-reinforcing bar is normally distributed with

Sample size,n = 49

Sample mean , X-bar = 8459 lb,

Standard deviations, σ = 100

We have to compute 90% CI for the true average yield point of the modified bar.

α = 1 - 90% = 1- 0.90 = 0.10

a)From Standard Normal Table, the critical value at the Zα/2 = Z₀.₀₅ level of significance is 1.645.

Confidence interval formula ,

X-bar +/- Zα/2(σ /√n)

90% CI for the true average yield point of the modified bar, is

8459 +/- 1.645(100/√49)

=> 8459 +/- 1.645(100/7)

=> 8459 +/- 1.645(14.2857) = 8459 +/- 23.5

b)

Now, we have to compute 98% CI for the true average yield point of the modified bar.

α = 1 - 98% = 1 - 0.98 = 0.02

From Standard Normal Table, the critical value at the Zα/2 = Z₀.₀₁ level of significance is 2.326.

Then, Confidence interval is,

CL = 8459 +/-2.326(100/√49)

= 8459 +/- 1.645( 14.285) = 8459 +/- 23.49

2) An article reported that for a sample of kitchens with gas cooking appliances monitored during a one-week period,

Sample size, n = 48

Sample mean , X-bar= 654.16

standard deviations, σ = 162.85

a) We have to calculate and interpret a 95% (two-sided) confidence interval for true average CO2 level. α = 1 - 0.95 = 0.05 ; α/2 = 0.025

From Standard Normal Table, the critical value at the Zα/2 = Z0.025 level of significance is 1.96

95% (two-sided) confidence interval for true average CO2 level is 654.16 +/- 1.96(654.16 /√48)

= 654.16 +/- 1.96(94.42)

= 654.16 +/- 185.0632

b) The distance between the two ends limits of the interval is known as the width of the interval and it is twice the margin of error.

We have width of the interval = 47 ppm

so, Margin of error , MOE = 94

Significance level, 95%

we have to determine sample size

Margin of error formula is

MOE = Zα/2(σ/√n)

where n is sample size.

94 = 1.96 (162.85/√n)

=> √n = 162.85 × 1.96/94 = 3.39559574468

=> n = 11.53 ~ 12

Hence, sample size is 12..

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The 90% confidence interval for the true mean proportion is given by

[-0.049 , -0.001]

Given values of the problem are:

P₁ = 12.9%

P₂ = 14.9%

n₁ = 1250

n₂ = 1000

hence the 90 % confidence interval can be calculated by using the formula for p-value and test statistic .

[tex]z = (P_1 - P_2 ) \pm z\times \sqrt{\frac{P_1\times(1-p_1)}{n_1}+\frac{P_2\times(1-{P_2})}{n_2}}[/tex]

Now we will put the above values in the equation to calculate the values of z .

Using the normal distribution table we can get the z-values.

Upper limit = - 0.049

Lower limit = - 0.001

Hence the 90% confidence interval can be calculated by [-0.049 , -0.001]

When employing common statistical procedures, confidence intervals are usually created using standardized techniques. The methods will have been created to fulfil a number of desired characteristics if the underlying presumptions are accurate. These desired properties include things like validity, optimality, and invariance.

Validity is the most important of the three, followed closely by "optimality." "Invariance" may be viewed as a characteristic of the procedure used to produce the confidence interval rather than a characteristic of the rule for generating the interval. In non-standard applications, the same desired attributes would be sought.

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The dimensions say (x,y,z) of a rectangular box of maximum volume is ( c/12, c/12, c/12).

The volume of a rectangular box is defined by the product of length, width, and height. For the maximum value of the rectangular box volume, simplify the expression by setting the first derivative of the rectangular box volume to zero in each dimension.

Let x, y, and z be the dimensions of the rectangular box. Since, the sum of the lengths of its 12 edges is a constant c. This implies

4x + 4y + 4z = c

=> x + y + z = c/4 --(1)

The volume of rectangular box = xyz

Consider xyz + lamda(x + y + z - c/4) where λ is Lagrange multiplier.

let f( x,y,z) = xyz + λ(x + y + z - c/4) --(*)

compute the partial derivatives of f( x,y,z) as

fx = yz + λ(1) = yz + λ

fy = xz + λ(1) = xz + λ

fz = xy + λ(1) = xy + λ

For critical points, set the partial derivatives equals to zero.

fx = yz + λ = 0

fy = xz + λ = 0

fz = xy + λ =0

=> xz = yz = xy = - λ

So, x = y = z

then, from equation (1) we get,

x + x + x = c/4

=> 3x = c/4 => x = c/12

thus, x = c/12, y = c/12, z = c/12

Thus, the dimensions of the rectangular box are x = y = z = c/12 or ( c/12, c/12, c/12) and

Maximum volume of rectangular box = xyz

= c³/1728

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Using the properties of the normal distribution we get the 90% confidence interval as (121.576 , 273.424)

The information given is : 185, 165, 85, 135, 80, and 225.

The mean and standard deviation can be determined.

Mean = x/n = 1185/6 = 197.5

46.125 is the standard deviation

The range of confidence

Error margin for the mean

T at 99%, df = n - 1; 6 - 1 = 5;

Margin of Error = Ts / √n

T = 4.032

And the margin of error is equal to 4.032 × 46.125 ÷ 6.

Error margin is 75.924

Range of confidence: 197.5 to 75.924

Lower border = 121.576 - 197.5 - 74.924

Upper limit: 197.5 + 74.924 = 273.424

is a collection of estimations for an unnamed parameter known as a confidence interval (CI). The most common confidence level is 95%, but when calculating confidence intervals, other levels, such 90% or 99%, are also occasionally employed.

The confidence level is a measure of how many related CIs over the long run include the actual value of the parameter. For instance, the parameter's true value should be included in 95% of all intervals produced at the 95% confidence level.

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the orthogonal projection of v onto the subspace W spanned by the vectors UI. v = 1 2 3 , u1 = 1 −1 1 , u2 = −1 1 2 then The orthogonal projection of v onto W is (1/6) (4, 4, 9).

To find the orthogonal projection of v onto W, we need to first find the orthogonal basis vectors for W. Since the vectors u1 and u2 are given to be orthogonal, we can use them as the basis vectors for W. We can then calculate the projection of v onto each of these vectors using the dot product. This gives us the components of the projection vector. Finally, we can multiply each component by the appropriate weighting factor to obtain the orthogonal projection vector.

Let v = (2, 3, 4)

Let u1 = (1, 0, 0) and u2 = (0, 1, 0)

The projection of v onto u1 is given by:

(2, 3, 4) • (1, 0, 0) = 2

The projection of v onto u2 is given by:

(2, 3, 4) • (0, 1, 0) = 3

The orthogonal projection vector of v onto W is then given by:

2u1 + 3u2 = (2, 3, 0)

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Answer:

  about 323.9 mph

Step-by-step explanation:

You want the speed of a plane that travels 750 miles in 2 hours with the wind and in 2.75 hours against the wind.

The relation between time, distance, and speed is ...

  speed = distance/time

If p represents the speed of the plane, and w represents the speed of the wind, then we have ...

  p +w = 750/2 . . . . . . miles/hour with the wind

  p -w = 750/2.75 . . . . speed against the wind

Adding these two equations gives ...

  2p = 750(1/2 +1/2.75) = 750(4.75/5.50)

Dividing by 2, we have ...

  p = 375(19/22) ≈ 323.9 . . . . . miles per hour

The plane was flying about 323.9 miles per hour.

__

Additional comment

The wind speed was about 51.1 miles per hour.

Answer:

20(3-2y)

Step-by-step explanation:

60−40y =

10(6-4y) =   ==> both 60 and 40 are multiples of 10

2(10(3-2y)) =  ==> both 6 and 4 are factors of 2

2*10(3-2y) = ==> simplify

20(3-2y)

The score does Dana need on the fourth examination to raise his average to 87 is 98.

What is the average of numbers?

Average by adding a group of numbers, dividing by their count, and then summing the results, the arithmetic mean is determined. For instance, the sum of 2, 3, 3, 5, 7, and 10 is equal to 30 divided by 6, which equals 5. Median the central number in a set of numbers.

Given: Dana received scores of  85, 92, and 73.

We have to find the score does Dana need on the fourth examination to raise his average to 87.

Suppose the score on the fourth examination is x.

The average of scores is 87.

⇒

[tex]87 = \frac{85 + 92 + 73 + x}{4} \\348 = 250 + x\\x = 348 - 250\\x = 98[/tex]

Hence, the score does Dana need on the fourth examination to raise his average to 87 is 98.

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Answer:

A) -18

B) 14

Step-by-step explanation:

A:

f(x) = 3x -6

f(-4) = 3(-4) - 6

f(-4) = -12 - 6

f(-4) = -18

B:

36 = 3x - 6  Add 6 to both sides

36 + 6 = 3x - 6 + 6

42 = 3x  Divide both side by 3

[tex]\frac{42}{3}[/tex] = [tex]\frac{3x}{3}[/tex]

14 = x

The Z-value of 2.58 implies a 97.72% confidence interval. This is because the Z-value of 2.58 corresponds to a probability of 0.9974 (which is 1 - 0.0026). This 0.9974 probability is equivalent to a 97.72% confidence interval.

The Z-value of 2.58 corresponds to a probability of 0.9974 (or 1 - 0.0026). To calculate this probability, we use the standard normal cumulative distribution function (CDF).

Using the standard normal CDF, we can calculate the cumulative probability of a Z-value of 2.58 by plugging in the Z-value into the equation and solving for the cumulative probability. The standard normal CDF is given by the following equation:

P(Z <= z) = 1 - 1/2*(1 + erf(z/sqrt(2)))

Plugging in the Z-value of 2.58 into the equation, we get:

P(Z <= 2.58) = 1 - 1/2*(1 + erf(2.58/sqrt(2)))

Solving for the cumulative probability, we get:

P(Z <= 2.58) = 0.9974

This 0.9974 probability is equivalent to a 97.72% confidence interval.

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The response variable was the weight of the package's standard deviation

a)the generator for this design is E=-ABCD

b)the resolution of this design is I=-ABCDE Therefor the resolution of the design is V.

c)estimate the factor effects there is 3 larger effcts namely E=-0.4700,BE=-0.4050,DE=-0.3150

d)Construct a linear regrassion model.The constant is estimated by the grand average and the regression coefficent are estimated by one-half the corresponding effect estimates.

If the underlying assumptions have any issues, the residual analysis will show them.

Y=1.22625+0.04375x₂-0.01875x₄+0.2350x₅-0.08125x₂x₄-0.1575x₄x₅

e)The standard deveation of package weight is affected the most by the dealy between mixing and packing,factor E.Also, its intreaction with the temperature factor B,BE and with the batch weight factor D,DE are important.

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The volume of the composite solid is equal to 72π cubic centimeters.

The volume of the composite solid shown in the figure is the result of the sum of the volumes of two solids: a cylinder and a cone. The volume formula of each element is shown below:

Volume of a cylinder

V = π · r² · h

Volume of a cone

V = (π / 3) · r² · h'

Where:

Volume of the composite solid

V = π · r² · h + (π / 3) · r² · h'

If we know that r = 3 cm, h = 7 cm and h' = 3 cm, then the volume of the composite solid is:

V = π · (3 cm)² · (7 cm) + (π / 3) · (3 cm)² · (3 cm)

V = 72π cm³

The composite solid has a volume of 72π cubic centimeters.

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(a) Sandy picked  [tex]1\frac{1}{2}[/tex] more buckets than Keith  .

(b) Jason have saved money for total of [tex]3\frac{3}{4}[/tex] months .

In the question ,,

Part(a)  ,

it is given that

number of buckets of apples Keith picked is =  [tex]2\frac{5}{6}[/tex] buckets ,

number of buckets of apples Sandy picked is = [tex]4\frac{1}{3}[/tex] buckets

number of more buckets picked by Sandy is = (buckets picked by Sandy) - (buckets picked by Keith) .

= [tex]4\frac{1}{3}[/tex]  -  [tex]2\frac{5}{6}[/tex]  

= 13/3 - 17/6  = 3/2 = [tex]1\frac{1}{2}[/tex] buckets .

Part(b)

number of months that Jason saved money in account = [tex]2\frac{1}{2}[/tex] months

number of months that Jason saved money in cash =  [tex]1\frac{1}{4}[/tex] months

total number of months for which money is saved is =  [tex]2\frac{1}{2}[/tex] +  [tex]1\frac{1}{4}[/tex]

Simplifying further ,

we get,

= 5/2 + 5/4

= 15/4

=  [tex]3\frac{3}{4}[/tex] months .

Therefore , (a) Sandy picked  [tex]1\frac{1}{2}[/tex] more buckets  and (b) Jason saved money for [tex]3\frac{3}{4}[/tex] months .

The given question is incomplete , the complete question is

(a) Keith picked [tex]2\frac{5}{6}[/tex] buckets of apples and Sandy picked [tex]4\frac{1}{3}[/tex] buckets. How many more buckets of apples did sandy pick ?

(b) Jayson has [tex]2\frac{1}{2}[/tex] months worth of pay saved in his account. He has [tex]1\frac{1}{4}[/tex] months worth of pay saved in cash. Altogether, how many months of  money has Jayson saved ?

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Answer:

He need to memorize 20 times tables.

Step-by-step explanation:

.2 x 100 = 20

The p-value (0.0324) < α (0.10), & There is sufficient proof that the best ski area is dependent on the level of the skier.

What is the p-value?

The p-value is the probability that we would see a test statistic this extreme or more if the null hypothesis were true.

(a) Hypotheses:          

H1 : Best ski area is independent of the level of the skier      

H2 : Best ski area is dependent on the level of the skier      

(b) Expected counts:          

                                              Beginner     Intermediate    Advanced     Total  

Tahoe     Observed                 20          30               40          90

             Expected               12.41            31.03            46.55   90.00

Utah     Observed                 10             30             60    100

             Expected              13.79           34.48           51.72  100.00

Colorado    Observed                 10             40             50    100

             Expected                13.79            34.48           51.72  100.00

Total     Observed                  40             100             150  290

             Expected               40.00          100.00       150.00     290.00

(c)

All the expected counts are > 5, so we can apply the test      

(d) Test statistic:

Square Contingency Table Test for Independence  

                                              Beginner     Intermediate    Advanced     Total

Tahoe     Observed                 20          30               40          90

             Expected               12.41            31.03            46.55   90.00

              O - E                       7.59            -1.03          -6.55   0.00

             (O - E)² / E               4.64              0.03           0.92   5.59

Utah     Observed                 10             30             60    100

             Expected              13.79           34.48           51.72  100.00

                      O - E               -3.79            -4.48          8.28   0.00

             (O - E)² / E               1.04              0.58           1.32    2.95

Colorado    Observed                 10             40             50    100

             Expected                13.79            34.48           51.72  100.00

                      O - E               -3.79            5.52          -1.72   0.00

             (O - E)² / E               1.04              0.88           0.06        1.98

Total     Observed                  40             100             150  290

             Expected               40.00          100.00       150.00     290.00

                      O - E               0.00            0.00          0.00    0.00

             (O - E)² / E               6.72               1.50            2.30        10.53

                                         10.53         chi-square    

                                                 4           df    

                                      0.0324           p-value  

x² = 10.53 and p-value = 0.0324        

(e) Since the p-value (0.0324) < α (0.10), we reject H1. There is sufficient evidence that the best ski area is dependent on the level of the skier.

Hence, the p-value (0.0324) < α (0.10), & There is sufficient proof that the best ski area is dependent on the level of the skier.

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The distribution is not normal. The points show a systematic pattern that is not a straight-line pattern curved over mean.

A normal distribution is a type of data distribution in which the data points form a symmetric, bell-shaped curve around the mean.

The data points in the example provided show a systematic pattern that is not a straight-line pattern, indicating that the data does not come from a normally distributed population.

This is because the points are not reasonably close to a straight line, which is a characteristic of a normal distribution.

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