a 14 ft long ladder is placed against a house with an angle of elevation of 72 degrees. How high above the ground is the top of the ladder?

Answer:

4.) Step 2; it should be m∠o + m∠p = 180 degrees (supplementary angles)

Step-by-step explanation:

o and p are supplementary angles, and therefore add up to 180 degrees.

Answer:

  A.  18.21 m/s  or  65.6 km/h

  B.  135.5 km/h

Step-by-step explanation:

A. The vertical speed of the grenade is ...

  vv = v·sin(45°) = v/√2

The time in air is given by the equation for ballistic motion:

  h = -4.9t² +vv·t = 0

  t(vv -4.9t) = 0 . . . . . factor out t

  v/√2 = 4.9t

  v = t·4.9√2 . . . . . . solve for v

__

The horizontal distance the grenade travels in t seconds is ...

  d = vh·t = v·cos(45°)·t = vt/√2 = (t·4.9√2)(t/√2) = 4.9t² . . . . meters

1 m/s = 3.6 km/h, so the horizontal distance the target travels is ...

  x = (109 - 81.0)/3.6·t + 13.4 = 70t/9 +13.4 . . . . meters in t seconds

We want the target distance to be the same as the grenade's distance, so the time required to hit the target is ...

  d = x

  4.9t² = (70/9)t +13.4

  4.9(t² -(100/63)t) = 13.4

  4.9(t -50/63)² = 13.4 +4.9(50/63)² ≈ 16.486

  t = 50/63 + √(16.486/4.9) ≈ 2.62793 . . . seconds

This corresponds to a launch speed of ...

  v = (4.9√2)t ≈ 18.21 . . . . meters/second

__

B. As we said, 1 m/s = 3.6 km/h, so the launch speed is about 65.558 km/h at an angle of 45° relative to the direction of travel. The magnitude of the velocity relative to the earth (ve) will be this vector added to 81.0 km/h in the direction of travel. The (horizontal, vertical) components of that sum are ...

  (81.0, 0) + 65.558(cos(45°), sin(45°)) ≈ (127.357, 46.357) km/h

The magnitude is the Pythagorean sum:

  ve = √(127.357² +46.357²) ≈ 135.5 . . . . km/h

The initial velocity Indy must throw the grenade at is found using the equations of motion, and it must be around 14.5 m/s. The magnitude of the velocity of the grenade relative to earth can be found by adding this initial velocity to the velocity of Indy's car, resulting in approximately 37 m/s.

This is a physics problem involving the concept of relative velocity. We know that when Indy lets go of the grenade, it will move at the same speed as his car (81.0 km/h) relative to the ground plus the additional speed he gives it. Let's assume Indy throws the grenade when the enemy car is exactly at this distance.

A. If the grenade is thrown at an angle of 45 degrees above the horizontal, we can apply the equations of motion to find the initial velocity of the grenade. Using the equation, Range = [v^2 * sin(2*angle)]/g, we can solve for v, which should be around 14.5 m/s.

B. The magnitude of the velocity of the grenade relative to earth would be the vector sum of the initial velocity of the grenade and the velocity of Indy's car. Converting the car's speed to m/s (81.0 km/h = 22.5 m/s) and using vector addition, we can add 22.5 m/s to the initial speed of the grenade (14.5 m/s) to receive a total of approximately 37 m/s.

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Answer:

(a) The p-value is 0.9554

(b) We cannot reject the null hypothesis at the significance level of 0.06

Step-by-step explanation:

We are dealing with a lower-tail alternative [tex]H_{1}: p < 0.4[/tex]. Because the test statistic is z = +1.7 which comes from a normal distribution, the p-value is the probability of getting a value as extreme as the already observed.  

(a) P(Z < +1.7) = 0.9554

(b) The p-value is very large, and 0.9554 > 0.06, so, we cannot reject the null hypothesis at the significance level of 0.06

Answer: a) 0.23, b) No, c) No.

Step-by-step explanation:

Since we have given that

Probability of homes for sale have garages = 61%

Probability of homes having swimming pool = 39%

Probability of homes having both = 14%

a) If a home for sale has a​ garage, what's the probability that it has a​ pool, too?

Using "Conditional probability", we get that

[tex]P(P|G)=\dfrac{P(P\cap G)}{P(G)}=\dfrac{0.14}{0.61}=0.23[/tex]

​b) Are having a garage and having a pool independent​ events? Explain.

P(P).P(S) = 0.61×0.39=0.24≠P(P∩S)=0.14

Hence, they are not independent events.

c) Are having a garage and having a pool mutually​ exclusive? Explain.

Since P(P∩S)≠0

So, they are not mutually exclusive.

Hence, a) 0.23, b) No, c) No.

Final answer:

To calculate the probability of a home having a pool given it has a garage, one uses the given percentages to find that it's approximately 23%. Examining whether having a garage and having a pool are independent or mutually exclusive reveals they are neither independent (because the probabilities differ when one condition is known) nor mutually exclusive (because homes can and do have both features).

Explanation:

Given information implies that 61% of homes have garages (G), 39% have pools (P), and 14% have both a garage and a pool (G∩P).

a) The probability that a home for sale has a pool given it has a garage can be calculated using the formula P(P|G) = P(G∩P) / P(G). Substituting the given values, we get P(P|G) = 0.14 / 0.61, which approximately equals 0.23 or 23%. This means if a home has a garage, there's a 23% chance it also has a pool.

b) Two events are independent if the occurrence of one does not affect the probability of the occurrence of the other. To check if G and P are independent, we calculate P(P) and compare it to P(P|G). Since P(P) = 0.39 and P(P|G) = 0.23, they are not equal; thus, having a garage and having a pool are not independent events. This is evident because knowing that a home has a garage changes the probability that it also has a pool.

c) Two events are mutually exclusive if they cannot occur at the same time. Given that 14% of homes have both a garage and a pool, it's clear that having a garage and having a pool are not mutually exclusive, as a home can have both features simultaneously.

Answer: No , at 0.05 level of significance , we have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

Step-by-step explanation:

Let [tex]\mu[/tex] denotes the average hours of sleep per night.

As per given , we have

[tex]H_0:\mu=7\\H_a:\mu<7[/tex]

, since [tex]H_a[/tex] is left-tailed and population standard deviation is unknown, so the test is a left-tailed t -test.

Also , it is given that ,

Sample size : n= 22

Sample mean : [tex]\overline{x}=7.24[/tex]

Sample  standard deviation : s= 1.93

Test statistic : [tex]t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]

i.e.  [tex]t=\dfrac{7.24-7}{\dfrac{1.93}{\sqrt{22}}}\approx0.58[/tex]

For significance level [tex]\alpha=0.05[/tex] and degree of freedom 21 (df=n-1),

Critical t-value for left-tailed test= [tex]t_{\alpha, df}=t_{0.05,21}=- 1.7207[/tex]

Decision : Since the test statistic value (0.58) > critical value  1.7207, it means we are failed to reject the null hypothesis .

[Note : When [tex]|t_{cal}|>|t_{cri}|[/tex], then we accept the null hypothesis.]

Conclusion: We have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

To determine if LTCC Intermediate Algebra students get less than seven hours of sleep per night, a one-sample t-test can be conducted.

To determine if LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average, we can conduct a one-sample t-test. The null hypothesis (H0) is that the mean number of hours of sleep is at least seven, while the alternative hypothesis (Ha) is that the mean number of hours of sleep is less than seven. With a significance level of 5%, we compare the t-statistic calculated from the sample data to the critical t-value from the t-distribution table. If the calculated t-statistic is less than the critical t-value, we reject the null hypothesis and conclude that LTCC Intermediate Algebra students get less than seven hours of sleep on average.

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To construct a confidence interval for the population mean with an unknown standard deviation, we use Student's t-distribution. The 99% confidence interval for the given sample mean (x = 880) and sample standard deviation (s = 5) is (877.371, 882.629). Assuming a different sample standard deviation, the 99% confidence intervals are (874.24, 885.76) for s = 10 and (868.48, 891.52) for s = 20. The confidence interval width increases as the sample standard deviation increases.

To construct a confidence interval for the population mean with an unknown standard deviation, we use Student's t-distribution. First, we find the value of the t-statistic for the given confidence level and degrees of freedom. For a 99% confidence level and a sample size of 24, the degrees of freedom is 23. Using Appendix D or a t-table, we find the critical t-value to be approximately 2.819.

(a) To construct a confidence interval with the given sample mean (x = 880) and sample standard deviation (s = 5), the margin of error is calculated as: E = t * (s / sqrt(n)) = 2.819 * (5 / sqrt(24)) = 2.819 * 1.021 = 2.88 (rounded to 3 decimal places). The 99% confidence interval is then calculated as: (x - E, x + E) = (880 - 2.88, 880 + 2.88) = (877.12, 882.88), rounded to 3 decimal places as (877.371, 882.629).

(b) Using the same method, but assuming a sample standard deviation of s = 10, the margin of error is calculated as: E = 2.819 * (10 / sqrt(24)) = 5.76 (rounded to 3 decimal places). The 99% confidence interval is then (874.24, 885.76).

(c) Assuming s = 20, the margin of error is calculated as: E = 2.819 * (20 / sqrt(24)) = 11.52 (rounded to 3 decimal places). The 99% confidence interval is then (868.48, 891.52).

(d) As the sample standard deviation (s) increases, the margin of error (E) and confidence interval width will also increase. This means that as s increases, we become less certain about the true population mean, resulting in a wider range of values in the confidence interval.

Answer:

∠NMP, so m∠1 = m∠2.

Step-by-step explanation:

Answer:

As x approaches 1 from the left, the value of [tex]arcsin(x)[/tex] is [tex]\frac{\pi }{2}[/tex]

Step-by-step explanation:

To find the value of [tex]arcsin(x)[/tex] when x approaches 1 from the left, you can use the graph of the function [tex]arcsin(x)[/tex]

Examine what happens as x approaches from the left.

As x approaches 1 from the left, the function seems to be approaching [tex]\frac{\pi }{2}[/tex]

Therefore, as x approaches 1 from the left, the value of [tex]arcsin(x)[/tex] is [tex]\frac{\pi }{2}[/tex]

As x approaches 1 from the left, the value of arcsin(x) approaches π/2 radians.

The expression x → 1− refers to the limit as x approaches 1 from the left. For the function presented, arcsin(x), we need to consider the value that this function approaches as x gets infinitely close to 1, but is still less than 1. This is asking what angle in radians has a sine of 1, or almost 1 from the left side. The sine of an angle is 1 when that angle is π/2 (or 90° in degree measure), hence as x → 1−, the value of arcsin(x) approaches π/2.

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Answer:

a) [tex]p[/tex] represent the real population proportion of people who showed partial or complete resistance to the antibiotic

b) [tex]\hat p=\frac{973}{1714}=0.568[/tex] represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

c) We are confident (95%) that the true proportion of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic is between 54.5% to 59.1%.  

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Part a

Description in words of the parameter p

[tex]p[/tex] represent the real population proportion of people who showed partial or complete resistance to the antibiotic

[tex]\hat p[/tex] represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

n=1714 is the sample size required

[tex]z_{\alpha/2}[/tex] represent the critical value for the margin of error

The population proportion have the following distribution  

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

Part b

Numerical estimate for p

In order to estimate a proportion we use this formula:

[tex]\hat p =\frac{X}{n}[/tex] where X represent the number of people with a characteristic and n the total sample size selected.

[tex]\hat p=\frac{973}{1714}=0.568[/tex] represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

Part c

The confidence interval for a proportion is given by this formula

[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.

[tex]z_{\alpha/2}=1.96[/tex]

And replacing into the confidence interval formula we got:

[tex]0.568 - 1.96 \sqrt{\frac{0.568(1-0.568)}{1714}}=0.545[/tex]

[tex]0.568 + 1.96 \sqrt{\frac{0.56(1-0.56)}{1714}}=0.591[/tex]

And the 95% confidence interval would be given (0.545;0.591).

We are confident that about 54.5% to 59.1% of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic.  

The scenario presents a population of strep-infected individuals in Florida with a parameter p representing the proportion resistant to penicillin. The estimate for this proportion is approximately 57%. The observation of resistance in a significant portion of the samples is consistent with growing concerns about antibiotic resistance.

The population in this case refers to all individuals in Florida who have been diagnosed with a strep infection. The parameter p here represents the proportion of these strep-infected individuals showing partial or complete resistance to the antibiotic penicillin.

The numerical value of the statistic that estimates p can be calculated by dividing the number of penicillin-resistant cultures by the total number of cultures. As a result, p = 973/1714 = 0.57 or approximately 57%.

For a 95% confidence interval for the proportion of strep cultures showing resistance to penicillin, we need to employ statistical methods. Without additional information, it would be impossible to calculate the confidence interval directly. However, it is important to remember that the true percentage of resistant cultures will likely lie within such an interval, which should be centered around the estimated proportion of 0.57 or 57%.

The observation that a significant portion of tested cultures show resistance to penicillin is a reminder of the growing concerns regarding antibiotic resistance. Overuse and misuse of antibiotics are believed to drive this development, with resistant bacteria such as MRSA (Methicillin-resistant Staphylococcus aureus) posing significant treatment challenges.

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Answer:

Length = 2 ft

Width = 2 ft

Height = 5 ft

Step-by-step explanation:

Let the square base of the box has one side = x ft

Therefore, area of the base = x² ft²

Cost of the material to prepare the base = $0.35 per square feet

Cost to prepare the base = $0.35x²

Let the height of the box = y ft

Then the volume of the box = x²y ft³ = 20

[tex]y=\frac{20}{x^{3} }[/tex] -----(1)

Cost of the material for the sides = $0.10 per square feet

Area of the sides = 4xy

Cost to prepare the sides of the box = $0.10 × 4xy

                                                             = $0.40xy

Cost of the material to prepare the top = $0.15 per square feet

Cost to prepare the top = $0.15x²

Total cost of the box = 0.35x² + 0.40xy + 0.15x²

From equation (1),

Total cost [tex]C=0.35x^{2}+(0.40x)\times \frac{20}{x^{2} }+0.15x^{2}[/tex]

[tex]C=0.35x^{2}+\frac{8}{x}+0.15x^{2}[/tex]

[tex]C=0.5x^{2}+\frac{8}{x}[/tex]

Now we take the derivative of C with respect to x and equate it to zero,

[tex]C'=0.5(2x)-\frac{8}{x^{2}}[/tex] = 0

[tex]x-\frac{8}{x^{2}}=0[/tex]

[tex]x=\frac{8}{x^{2} }[/tex]

[tex]x^{3}=8[/tex]

x = 2 ft.

From equation (1),

[tex](2)^{2}y=20[/tex]

4y = 20

y = 5 ft

Therefore, Length and width of the box should be 2 ft and height of the box should be 5 ft for the minimum cost to construct the rectangular box.

The volume of a box is the amount of space in it.

The dimensions that minimize cost are:

[tex]\mathbf{Length = 2ft}[/tex]

[tex]\mathbf{Width = 2ft}[/tex]

[tex]\mathbf{Height= 5ft}[/tex]

The volume of the box is:

[tex]\mathbf{V = 20}[/tex]

Assume the dimension of the base is x, and the height is h.

The volume of the box will be:

[tex]\mathbf{V = x^2h}[/tex]

So, we have:

[tex]\mathbf{x^2h = 20}[/tex]

The area of the sides is:

[tex]\mathbf{A_1 = 4xh}[/tex]

The cost of the side material is 0.10.

So, the cost is:

[tex]\mathbf{C_1 = 0.10 \times 4xh}[/tex]

[tex]\mathbf{C_1 = 0.4xh}[/tex]

The area of the base is:

[tex]\mathbf{A_2 = x^2}[/tex]

The cost of the base material is 0.35.

So, the cost is:

[tex]\mathbf{C_2 = 0.35x^2}[/tex]

The area of the top is:

[tex]\mathbf{A_3 = x^2}[/tex]

The cost of the top material is 0.15.

So, the cost is:

[tex]\mathbf{C_3 = 0.15x^2}[/tex]

So, the total cost is:

[tex]\mathbf{C = 0.4xh + 0.35x^2 + 0.15x^2}[/tex]

[tex]\mathbf{C = 0.4xh + 0.5x^2}[/tex]

Recall that: [tex]\mathbf{x^2h = 20}[/tex]

Make h the subject

[tex]\mathbf{h = \frac{20}{x^2}}[/tex]

Substitute [tex]\mathbf{h = \frac{20}{x^2}}[/tex] in [tex]\mathbf{C = 0.4xh + 0.5x^2}[/tex]

[tex]\mathbf{C = 0.4x \times \frac{20}{x^2} + 0.5x^2}[/tex]

[tex]\mathbf{C = \frac{8}{x} + 0.5x^2}[/tex]

Differentiate

[tex]\mathbf{C' = -\frac{8}{x^2} + x}[/tex]

Set to 0

[tex]\mathbf{-\frac{8}{x^2} + x = 0}[/tex]

Rewrite as:

[tex]\mathbf{x = \frac{8}{x^2}}[/tex]

Multiply both sides by x^2

[tex]\mathbf{x^3 = 8}[/tex]

Take cube roots of both sides

[tex]\mathbf{x = 2}[/tex]

Recall that:

[tex]\mathbf{h = \frac{20}{x^2}}[/tex]

So, we have:

[tex]\mathbf{h = \frac{20}{2^2}}[/tex]

[tex]\mathbf{h = 5}[/tex]

Hence, the dimensions that minimize cost are:

[tex]\mathbf{Length = 2ft}[/tex]

[tex]\mathbf{Width = 2ft}[/tex]

[tex]\mathbf{Height= 5ft}[/tex]

Read more about volumes at:

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Answer:

Step-by-step explanation:

Hello!

Remember, the rule to decide using the p-value is always the same.

If the p-value ≤ α, you reject the null hypothesis.

If the p-value > α, you support the null hypothesis.

The experiment states the hypothesis that the proportion of babies born with low weight is higher if their mothers were exposed to second-hand smoking during pregnancy. Symbolically:

H₀: ρ ≤ 0.078

H₁: ρ > 0.078

α: 0.05

The given p-value is 0.119

Since the p-value (0.119) is greater than the level of significance (α: 0.05) the decision is to not reject the null hypothesis.

The proportion of babies born with low weight among women that were exposed to second-hand smoking during pregnancy is no different from the proportion in the general population.

I hope it helps!

Answer:

Z= -1.429

Step-by-step explanation:

Hello!

The owner thinks that the number of customers decreased. The hystorical average per day is 800 customers.

So the variable X: " Customers per day" X~N(μ;σ²)

σ= 350

Symbolically the test hypothesis are:

H₀: μ ≥ 800

H₁: μ < 800

The statistic value is:

Since you are studying the population sample, the study variable has a normal distribution and you know the value of the population variance, the propper statistic to make the test is:

Z= X[bar] - μ = 750 - 800 = -1.4285 ≅ -1.429

     σ/√n           350/√100

where X[bar] is the sample mean

μ is the population mean under the null hypothesis

σ is the population standard  

n is the sample taken

The value is Z= -1.429

I hope it helps!

Answer:

the answer D=1.44

Step-by-step explanation:

if the X= number of red cards observed in 6 trials , since each card observation is independent from the others and the sampling process is done with replacement ( the card is observed, then returned and reshuffled) , X follows an binomial distribution.

X(x)= n!/((n-x)!*x!) *p^x *(1-p)^(n-x)

where n = number of trials = 6 , x= number of red cards observed , p= probability of obtaining a red card in one try

the probability of obtaining the card in one try is

p = number of red cards / total number of cards = 12/ (12+8) = 0.6

since we know that X has a binomial distribution, the variance of this kind of distribution is

variance = σ² = n * p * (1-p)

therefore the variance of X is

variance = σ² = n * p * (1-p) = 6 * 0.6 * (1-0.6) = 1.44

Answer:

a) Null hypothesis:[tex]\mu_{s}-\mu_{n}\geq 10[/tex]

Alternative hypothesis:[tex]\mu_{s} - \mu_{n}<10[/tex]

[tex]p_v =P(Z<-1.904)=0.0284[/tex]

Comparing the p value with the significance level [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly lower than 10.

b) The 90% confidence interval would be given by [tex]-21.035 \leq \mu_1 -\mu_2 \leq 7.685[/tex]

c) If we analyze the interval obtained we see that the interval contains the value of -10 so then we agree with the result obtained from the hypothesis, that he alternative hypothesis is true.

Step-by-step explanation:

Data given and notation

[tex]\bar X_{s}=750.2[/tex] represent the mean for the sample standard  process

[tex]\bar X_{n}=756.875[/tex] represent the mean for the sample with the new process

[tex]s_{s}=19.128[/tex] represent the sample standard deviation for the sample standard process

[tex]s_{n}=21.283[/tex] represent the sample standard deviation for the new process

[tex]\sigma_s=\sigma_n=\sigma=20[/tex] represent the population standard deviation for both samples.

[tex]n_{s}=15[/tex] sample size for the group Cincinnati

[tex]n_{n}=8[/tex] sample size for the group Pittsburgh

z would represent the statistic (variable of interest)

Concepts and formulas to use

(a) Formulate and test an appropriate hypothesis using a 0.10. What are your conclusions?

We need to conduct a hypothesis in order to check if the difference in mean batch viscosity is 10 or less system of hypothesis would be:

Null hypothesis:[tex]\mu_{s}-\mu_{n}\geq 10[/tex]

Alternative hypothesis:[tex]\mu_{s} - \mu_{n}<10[/tex]

We have the population standard deviation, so for this case is better apply a z test to compare means, and the statistic is given by:

[tex]z=\frac{(\bar X_{s}-\bar X_{n})-\Delta}{\sqrt{\frac{\sigma^2_{s}}{n_{s}}+\frac{\sigma^2_{n}}{n_{n}}}}[/tex] (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

With the info given we can replace in formula (1) like this:

[tex]z=\frac{(750.2-756.875)-10}{\sqrt{\frac{20^2}{15}+\frac{20^2}{8}}}}=-1.904[/tex]  

Statistical decision

Since is a one tail left test the p value would be:

[tex]p_v =P(Z<-1.904)=0.0284[/tex]

Comparing the p value with the significance level [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly lower than 10.

(b) Find a 90% confidence interval on the difference in mean batch viscosity resulting from the process change.

The confidence interval for the difference of means is given by the following formula:  

[tex](\bar X_s -\bar X_n) \pm z_{\alpha/2}\sqrt{\sigma^2(\frac{1}{n_s}+\frac{1}{n_s})}[/tex] (1)  

The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:

[tex]\bar X_s -\bar X_n =750.2-756.875=-6.675[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]  

The standard error is given by the following formula:

[tex]SE=\sqrt{\sigma^2(\frac{1}{n_s}+\frac{1}{n_n})}[/tex]

And replacing we have:

[tex]SE=\sqrt{20^2(\frac{1}{15}+\frac{1}{8})}=8.756[/tex]

Confidence interval

Now we have everything in order to replace into formula (1):  

[tex]-6.675-1.64\sqrt{20^2(\frac{1}{15}+\frac{1}{8})}=-21.035[/tex]  

[tex]-6.675+1.64\sqrt{20^2(\frac{1}{15}+\frac{1}{8})}=7.685[/tex]  

So on this case the 90% confidence interval would be given by [tex]-21.035 \leq \mu_1 -\mu_2 \leq 7.685[/tex]

(c) Compare the results of parts (a) and (b) and discuss your findings.

If we analyze the interval obtained we see that the interval contains the value of -10 so then we agree with the result obtained from the hypothesis, that he alternative hypothesis is true.

Final answer:

The viscosity analysis involves a hypothesis test to check for significant changes in mean batch viscosity and constructing a confidence interval to estimate the range of this difference following a process change in the manufacturing of a polymer.

Explanation:

The question involves conducting a hypothesis test and constructing a confidence interval to analyze the change in mean batch viscosity of a polymer before and after a process change. A hypothesis test will be used to determine if the mean batch viscosity has changed significantly, and the confidence interval will provide a range in which the true difference in means likely falls.

Hypothesis Test:

Calculate the mean of both sets of viscosity measurements.

Set up the null hypothesis (H0) and the alternative hypothesis (H1): H0: μ1 - μ2 <= 10 (no significant change), H1: μ1 - μ2 > 10 (significant change).

Determine the test statistic using a t-test for two independent samples.

Calculate the p-value associated with the test statistic.

Compare the p-value with the significance level (α = 0.10). If p <= α, reject H0; otherwise, fail to reject H0.

Calculate the standard deviation of both sets of measurements.

Determine the standard error of the difference in means.

Find the appropriate t-distribution critical value based on the given confidence level (90%) and degrees of freedom.

Calculate the 90% confidence interval using the difference in means ± the margin of error.

Compare the results of the hypothesis test and the confidence interval. If the hypothesis test leads to rejection of the null hypothesis while the confidence interval for the difference in means includes values greater than 10, there is evidence suggesting a significant change in mean batch viscosity. However, if the hypothesis test does not lead to rejection of H0 and the confidence interval includes 10, it suggests that the process change did not have a significant effect on mean batch viscosity.

Answer:

A) As α increases, β decreases

Step-by-step explanation:

I am assuming that the probabilities of both kind of error are positive, otherwise there is no reason to make an hypothesis test.

Both events arent possible simultaneously, so they cant be independent because the probability of both kind of errors is greater than 0.

If the p-value increases, then by definition, the probability of type 1 error increases, in this case, you are more likely to reject the null hypothesis, therefore you are less likely to make the mistake of not rejecting the null hypothesis when you have to. Thus, the probability of type 2 error decreases.

On the other way around, if you are less likely to reject the null hypothesis (hence, the probability of type 1 error decreases), then you will be more likely to not reject the null hypothesis even on weird scenarios, therefore the probability of type 2 error increases.

There is no formula to determine the relatioship between both errors, but the errors are related: when the probability of type 1 error increases, then the probability of type 2 error decreases. The correct answer is A)

Answer:

The forecast error for the decomposition method was within ±3% and the multiple regression error was also within ±3%. Therefore, both methods provide low error and similar forecast accuracy.

Step-by-step explanation:

January sales for the fourth year as stated  is $295,000.

While predicted sales  of decomposition method and forecasted sales of multiple regression method were $298,424 and $286,736  respectively.

Our assumption is  that for the accuracy to be valid, there is need for  it needs to be within ±3% area of the actual results .

Therefore:

295000 × 0.03 = 8850

295000 + 8850 = 303850  ( expected predicted sales of decomposition by ±3%)

295000 - 8850 = 286150   (expected forecasted sales for multiple regression  by ±3%)

We could see that both the results are within this ±3% area and decomposition was slightly more accurate. The forecast error for the decomposition method was within ±3% and the multiple regression error was also within ±3%. Therefore, both methods provide low error and similar forecast accuracy.

That's it!

Answer:

Independent Sampling

Step-by-step explanation:

There are two scenarios for independent sampling .

Testing the mean we get the differences between samples from each population. When both samples are randomly  inferences about the populations. can get.

Independent sampling are sample that are selected randomly. Observation does not depend upon value.Many analysis assume that sample are independent.

In this statement 90 dash are divides into two groups Group 1 and Group 2 . Both are standardized that mean both are randomly selected. Means are observed. Observation doesn't depend upon value.  So this style of sampling is independent Sample.

Answer:

We do not have enough evidence to accept H₀

Step-by-step explanation:

Normal Distribution

size sample  =  n = 64   (very small sample for evaluating population of 5  years

Standard deviation 4,8

1.- Test hypothesis

H₀             null hypothesis        ⇒               μ₀ = 14       and

Hₐ     alternative hypothesis   ⇒                μ₀ ≠ 14

2.- z(c)  we assume α = 0,05  as we are dealing with a two test tail we should  consider   α/2  = 0.025.

From z table we the z(c) value

z(c) = 1.96           and of course by symmetry   z(c) = -1.96

3.- We proceed to compute z(s)

z(s)  = [ (  μ -  μ₀ ) /( σ/√n) ]           ⇒    z(s)  = - (1.5)*√64/4.8

z(s)  = - 2.5

We compare z(s)  and z(c)

z(s) < z(c)     -2.5  < -1.96  meaning  z(s) is in the rejection zone

we reject H₀ .

From the start we indicate sample size as to small for the experiment nonetheless we found that we dont have enough evidence to accept H₀

Answer:

2.5

Step-by-step explanation:

We are given that

Mean=[tex]\mu=3.2[/tex] pounds

Standard deviation=[tex]\sigma=0.8[/tex] pounds

n=25

We have to find the Z-score if the mean of  a sample of 25 roasts is 3.6 pounds.

We know that Z-score formula

[tex]Z=\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

We have [tex]\bar X=3.6[/tex]

Substitute the values then we get

Z- score=[tex]\frac{3.6-3.2}{\frac{0.8}{\sqrt{25}}}[/tex]

Z- score=[tex]\frac{0.4}{\frac{0.8}{5}}=\frac{0.4\times 5}{0.8}=2.5[/tex]

Hence, the Z-score value for the given sample mean=2.5

The Z-score for a sample mean of 3.6 pounds for 25 roasts, given a population mean of 3.2 pounds and a population standard deviation of 0.8 pounds, is calculated through the formula 'Z = (X - μ) / (σ / √n)' and is determined to be 2.5. The fact that the Z-score is positive indicates that the sample mean is above the population mean.

The question presented involves calculation of a Z-score in a situation involving weights of roasts. The Z-score is a measure used in statistics that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. Here, we can use the formula for calculating Z-score from a sample mean:

Z = (X - μ) / (σ / √n)

where 'X' is the sample mean, 'μ' is the population mean, 'σ' is the population standard deviation, and 'n' is the size of the sample. Plugging in the values given in the problem, we get:

Z = (3.6 - 3.2) / (0.8 / √25) = 0.4 / 0.16 = 2.5

So, the Z-score for the sample mean of 25 roasts weighing 3.6 pounds each is 2.5.

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Answer:

d=0.0167mille

Step-by-step explanation:

we must determine its position (X-Y).

ship 1

[tex]v_{1}=8\frac{m}{h}\\t_{1}=12.5 h\\t_{2}=14h\\t_{t2}=t_{2}-t_{1}=14-12.5=1.5h\\d_{s1}=v_{1}*t_{t1}=8\frac{m}{h}*1.5h=12m[/tex]

ship 2

[tex]v_{2}=10\frac{m}{h}\\t_{1}=12.5 h\\t_{2}=14h\\t_{t2}=t_{2}-t_{1}=14-13.5=0.5h\\d_{s2}=v_{1}*t_{t2}=10\frac{m}{h}*0.5h=5m[/tex]

component  x-y

[tex]sin55=\frac{y_{1}}{h};y_{1}=12*sin55=9.82\\   cos55=\frac{x_{1} }{h};x_{1}=5*cos55=6,88[/tex]

[tex]sin150=\frac{y_{2}}{h};y_{2}=5*sin150=2.5\\cos150=\frac{x_{2} }{h};x_{2}=5*cos150=-4.33[/tex]

[tex]s_{1}=(6.88,9.82); s_{2}=(-4.33,2,5)[/tex]

we must find the distance S1-S2

[tex]d_{s1-s2}=\sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}}=\sqrt{(2.5-9.8)^{2}+(-4.33-6.88)^{2}}\\=\sqrt{(24.5)^{2}+(-11.21)^{2}}=26.94m[/tex]

but the units are requested to be miles

d=26.94m*mille/1,609.34m=0.0167mille

The probability of drawing a spade from a 52-card deck is 0.25. If it's known that the card is black, the probability that it's a spade increases to 0.5.

The subject of this question is the calculation of probability in a card game scenario. When a single card is selected from a standard 52-card deck, the probability that the card drawn is a spade​ is calculated by dividing the total number of spades in the deck (13) by the total number of cards in the deck (52). This gives a probability of 13/52 or 0.25.

However, if a single card is drawn from a standard 52-card deck, and it is known that the card is black, then the total number of possible outcomes is reduced to 26 (the number of black cards in the deck). In this scenario, the probability that the card drawn is a spade becomes 13/26 or 0.5.

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The probability of drawing a spade from a 52-card deck is 1/4. If we already know the card is black, the probability increases to 1/2.

This question relates to the field of probability. To address part a, in a standard 52-card deck, there are 13 spades. We therefore have a 13 out of 52 chance to draw a spade, simplifying to a probability of 1/4.

For part b, being told that the card drawn is black means our sample space becomes 26 (13 spades and 13 clubs). Since there are still 13 spades, the probability of drawing a spade given that the card is black is 13 out of 26, or 1/2.

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Answer:

we reject H₀

Step-by-step explanation:

Normal Distribution

sample size      n  =  25        degees of fredom     =  25 - 1  df = 24

sample standard deviation = s = 0,24

sample mean     11.88

We have a one tail test (left) investigation

1.-Test hypothesis

H₀          ⇒              null hypothesis         μ₀  =  12

Hₐ          ⇒     Alternative hypothesis     μ₀  <  12  

2.-Significance level   0,05     t(c)  = - 1.7109

3.-Compute of t(s)

t(s)  = (  μ  -   μ₀ )/s/√n          ⇒  t(s)  =[ ( 11.88 - 12 )*√25 ]/0.24

t(s)  = - 0.12*5/0.24

t(s)  = - 2.5

4.-We compare  t(s) with  t(c)

   In this case   t(s) < t(c)        - 2.5  <  -1.71

5.-t(s) is in the rejection region, we reject H₀

The machine is not adjusted

Answer:

[tex]D_{up}(t)=3t[/tex]

[tex]D_{down}(t)=8t[/tex]

[tex]D_{flat}(t)=5t[/tex]

Explanation:

To construct a linear model of their distance, in miles, as a function of time in hours, since we were given their traveling speed in miles per hour, simply multiply the traveling speed in mph by the time, t, in hours.

When traveling up a hill:

[tex]D_{up}(t)=3t[/tex]

When traveling down a hill:

[tex]D_{down}(t)=8t[/tex]

When traveling along a flat portion:

[tex]D_{flat}(t)=5t[/tex]

Final answer:

To construct linear models that describe the distance traveled on different portions of the path in terms of time, we can use the formula for distance: D = V * T, where D is the distance, V is the velocity, and T is the time.

Explanation:

To construct linear models that describe the distance traveled on different portions of the path in terms of time, we can use the formula for distance:

D = V * T

where D is the distance, V is the velocity, and T is the time.

For the uphill portion, the speed is 3 mph, so the linear model would be:

D = 3T

For the downhill portion, the speed is 8 mph, so the linear model would be:

D = 8T

For the flat portion, the speed is 5 mph, so the linear model would be:

D = 5T

Answer:

There is no sufficient evidence to reject the company's claim at the significance level of 0.05

Step-by-step explanation:

Let [tex]\mu[/tex] be the true mean weight per apple the company ship.  We want to test the next hypothesis

[tex]H_{0}:\mu=120[/tex] vs [tex]H_{1}:\mu\neq 120[/tex] (two-tailed test).

Because we have a large sample of size n = 49 apples randomly selected from a shipment, the test statistic is given by

[tex]Z=\frac{\bar{X}-120}{\sigma/\sqrt{n}}[/tex] which is normally distributed. The observed value is  

[tex]z_{0}=\frac{122.5-120}{12/\sqrt{49}}=1.4583[/tex]. The rejection region for [tex]\alpha = 0.05[/tex] is given by RR = {z| z < -1.96 or z > 1.96} where the area below -1.96 and under the standard normal density is 0.025; and the area above 1.96 and under the standard normal density is 0.025 as well. Because the observed value 1.4583 does not fall inside the rejection region RR, we fail to reject the null hypothesis.

To determine whether there is sufficient evidence to reject the company's claim, we can perform a hypothesis test.

To determine whether there is sufficient evidence to reject the company's claim, we can perform a hypothesis test. The null hypothesis, H0, is that the mean weight per apple is 120 grams, and the alternative hypothesis, Ha, is that the mean weight per apple is not 120 grams. With a sample of 49 apples, we can calculate the test statistic using the formula:

t = (sample mean - population mean) / (standard deviation / sqrt(sample size))

Once we have the test statistic, we can compare it to the critical value from the t-distribution to determine if we reject or fail to reject the null hypothesis. As per calculation, yhe area below -1.96 and under the standard normal density is 0.025; and the area above 1.96 and under the standard normal density is 0.025 as well.

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Answer:

GCF = 8

Step-by-step explanation:

Find the prime factorization of 48

48 = 2×2×2×2×3

Find the prime factorization of 56

56 = 2×2×2×7

To find the GFC, multiply all the prime factors common to both numbers

Therefore, GCF = 2×2×2

GCF = 8

The greatest common factor (GCF) of 48 and 56 is B. 8.

The greatest common factor (GCF) of two or more numbers is the largest number that divides evenly into all of the given numbers. To find the GCF of 48 and 56, we need to find the largest number that is a factor of both 48 and 56.

One method to find the GCF is to list the factors of each number and find the largest common factor. The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48. The factors of 56 are 1, 2, 4, 7, 8, 14, 28, and 56. By comparing the lists, we can see that the largest common factor is 8.

Another method to find the GCF is to use prime factorization. Prime factorization involves breaking down each number into its prime factors and finding the common prime factors. The prime factorization of 48 is 2^4 * 3, and the prime factorization of 56 is 2^3 * 7. To find the GCF, we take the product of the common prime factors with the lowest exponents, which in this case is 2^3, resulting in 8.

In both methods, we find that the GCF of 48 and 56 is 8. Therefore, the correct answer is B. 8.

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Answer:

99.8% confidence:

[0.6833, 0.7656]

99% confidence:

0.6902 < p < 0.7588

Step-by-step explanation:

Lets call p the probability that a U.S adult pay its credit catrd in full each month. Lets call Y the random variable that counts the total of persons that paid their credit card in full from a random sample of 1118 adults. Y is a random variable with distribution Y ≈ Bi(1118,p) . The expected value of Y is μ = 1118*p and its variance is σ² = 1118*p(1-p). The proportion of adults that paid their credit card is a random variable, lets call it X, obtained from Y by dividing by 1118. The expected value is p and the variance is p(1-p)/1118. The Central Limit Theorem states that X can be approximated by a random variable with Normal Distribution, with mean μ = p, and standard deviation σ = √(p(1-p)/1118).

The standarization of X, W, is a random variable with distribution (approximately) N(0,1) obtained from X by substracting μ and dividing by σ. Thus

[tex] W = \frac{X - \mu}{\sigma} = \frac{X - p}{\sqrt{\frac{p(1-p)}{1118}}} [/tex]

If we want a 99.8% confidence interval, then we can find a value Z such that P(-Z < W < Z) = 0.998. If we do so, then P(W < Z) = 0.999, therefore, Ф(Z) = 0.999, were Ф is the cummuative distribution function of the standard normal distribution. The values of Ф can be found on the attached file. We can find that Ф(3.08) = 0.999, thus, Z = 3.08.

We have that P(-3.08 < W < 3.08) = 0.998, in other words

[tex] P(-3.08 < \frac{X-p}{\sqrt{\frac{p(1-p)}{1118}}} < 3.08) = 0.998 [/tex]

[tex] P(-3.08 * \sqrt{\frac{p(1-p)}{1118}} < X-p < 3.08 * \sqrt{\frac{p(1-p)}{1118}}) = 0.998 [/tex]

[tex] P(-X -3.08 * \sqrt{\frac{p(1-p)}{1118}} < -p < -X + 3.08 * \sqrt{\frac{p(1-p)}{1118}}) = 0.998 [/tex]

Taking out the - sign from the -p and reversing the inequalities, we finally obtain

[tex] P(X -3.08 * \sqrt{\frac{p(1-p)}{1118}} < p < X + 3.08 * \sqrt{\frac{p(1-p)}{1118}}) = 0.998 [/tex]

As a conclusion, replacing p by its approximation X, a 99.8% confidence interval for p is

[tex] [X -3.08 * \sqrt{\frac{X(1-X)}{1118}}\, ,  \, X + 3.08 * \sqrt{\frac{X(1-X)}{1118}}] [/tex]

replacing X with the proportion of the sample, 810/1118 = 0.7245, we have that our confidence interval is

[tex] [0.7245 -3.08 * \sqrt{\frac{0.7245(1-0.7245)}{1118}}\, ,  \, 0.7245 + 3.08 * \sqrt{\frac{0.7245(1-0.7245)}{1118}}] [/tex]

By solving the fraction and multypling by 3.08, we have

[0.7245 - 0.0411 < p < 0.7245 + 0.0411]

FInally, the 99.8 % confidence interval for p is

[0.6833, 0.7656]

If p is 0.99 (99% confidence), then we would want Z such that Ф(Z) = 0.995, by looking at the table we have that Z is 2.57, therefore the 99% interval for p is

[tex] [X -2.57 * \sqrt{\frac{X(1-X)}{1118}}\, ,  \, X + 2.57 * \sqrt{\frac{X(1-X)}{1118}}] [/tex]

and, by replacing X by 0.7245 we have that  the 99% confidence interval is

[0.6902, 0.7588]

thus, 0.6902 < p < 0.7588

I hope i could help you!

Final answer:

To construct a 99.8% confidence interval for the proportion of U.S. adults who pay their credit cards in full, we use the sample proportion, the z-value for the confidence level, and the sample size to calculate the margin of error and produce the interval.

Explanation:

To construct a 99.8% confidence interval for the proportion of U.S. adults who pay their credit cards in full each month from a survey result where 810 out of 1118 respondents paid their credit cards in full, we will use the formula for the confidence interval of a proportion: CI = p ± Z*sqrt((p(1-p))/n). Here, 'p' is the sample proportion, 'Z' is the z-value corresponding to the desired confidence level, and 'n' is the sample size.

The sample proportion (p) is 810/1118. To find the appropriate 'Z' value for a 99.8% confidence interval, we can refer to a standard normal distribution table or use a calculator to find that Z is approximately 3.08.

Step-by-step:

Find the sample proportion: p = 810/1118

Determine the 'Z' value for 99.8% confidence level: Z ≈ 3.08

Calculate the standard error (SE): SE = sqrt(p(1-p)/n)

Calculate the margin of error (ME): ME = Z * SE

Construct the confidence interval: (p - ME, p + ME)

After calculations, we can state that the confidence interval for the proportion of U.S. adults who pay their credit cards in full each month is between two values with three decimal places precision (the exact numbers would need to be calculated).

Answer:

36x^2

Step-by-step explanation:

Since area is base*height:

8*4.5=36

x*x=x^2

The monomial is 36x^2

Answer:

True

Step-by-step explanation:

The answer is "True" because the data set was picked at random.

Answer:

0.546 , -4.71

Step-by-step explanation:

Given:

An angle's initial ray points in the 3-o'clock direction and its terminal ray rotates counter -clock wise.

Here, Slope = tan\theta

If θ = 0.5

Then, Slope = tan(θ) = tan(0.5) = 0.546

If θ = 1.78

Then, Slope = tan(θ) = tan(1.78) = - 4.71

The expression (in terms of θ) that represents the varying slope of the terminal ray.

Slope = m = tanθ, where θ is the varying angle

A) The slope of the terminal ray when θ = 0.5 radians is; 0.5463

B) The slope of the terminal ray when θ = 1.78 radians is; -4.71

C) The expression that will represent the varying slope of the terminal ray is;

tan (Δy/Δx)

We are given that θ represents the angle's varying measure (in radians).

Now, in mathematics, slope is simply the tangent of an angle. Thus;

A) At θ = 0.5 radians ,

Slope of terminal ray = tan θ

Slope = tan 0.5

Using radians calculator, tan 0.5 = 0.5463

Thus, slope = 0.5463

B) At θ = 1.78

Slope of terminal ray = tan θ

Slope = tan 1.78

Using radians calculator, tan 1.78 = -4.71

Thus, slope = -4.71

C) The expression that will represent the varying slope of the terminal ray is;

tan (Δy/Δx)

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