Nuri joins a game for a car. The rule is that Nuri pick one key from box either A, B, or C. A box has two keys but only one can be used. B box has three keys but only one can be used. C box has two keys but none of them can be used. What is the probability that Nuri can win the car?

Answer:

x (max) = 26760 m

y (max)  = 3859  meters

V = 549.5  m/sec

Step-by-step explanation:

Equations to describe the projectile shot movement are:

a(x)  =  0     V(x)  =  V(₀) *cos α                  x  =  V(₀) *cos α  * t

a(y)  = -g     V(y)  =  V(₀) * sin α  - g*t         y  =   V(₀) * sin α *t  - (1/2)*g*t²

a ) What is the range of the projectile.   α  =  30°

then  sin 30° = 1/2     cos  30° = √3 /2   and     tan 30° = 1/√3

x  maximum occurs when in the equation of trajectory  we make  y  = 0

Then

y  =  x*tan α  -  g*x / 2*V(₀)²*cos²  α

x*tan α  =  g*x /  2* V(₀)²*cos²  α

By subtitution

1/√3    =   9.8* x(max)  / 2* (550)²*0.75

(1/√3) * 453750 / 9.8  = x (max)

x (max) = 453750 / 16.95   meters

x (max) = 26760 m

The maximum height is when V(y) = 0

We compute t in that condition

V(y)  =  0  =  V(₀) * sin α - g*t

t  =   V(₀) * sin α / g       ⇒   t  =  550* (1/2) / 9.8

t  =  28.06 sec

Then  h (max)  =   y(max)  =  V(₀) sin α * t  - 1/2 g* t²

y (max)  =  550* (1/2)*28.06 - (1/2)* 9.8 * (28.06)²

y (max)  =  7717  -  3858  

y (max)  = 3859  meters

What is the speed when the projectile hits the ground

V  =  V(x)  + V (y)    and   t  =  2* 28.06         t  =  56.12 sec

mod V  =√ V(x)²  + V(y)²

V(x)  =  V(₀) cos α    =  550 * √3/2

V(x)  = 475.5  m/sec       V(x)²   =  226338 m²/sec²

V(y) =  550*1/2   -  9.8* 56.12     ⇒  V(y) = 275  -  549.98

V(y) =  - 274.98          V(y) ²   =  

V =  √ 226338 + 75614     ⇒  V = 549.5  m/sec

They put 20% down so that means 80% was financed ( 100% - 20% = 80%).

Multiply the purchase price by 80%:

23750 x 0.80 = $19,000 was financed.

Answer:

6 hours

Step-by-step explanation:

You can set up a chart:

      R       T     D

O  300   1+x   ~                  O=out   B=back R=Rate/Speed T=time D=distance

B   350   x      ~

The distance is the same. On the way out the speed was 300 mph, so you can see where that is on the chart. And the speed back was 350 mph. It took an extra hour on the way out than on the way back, so you can call the way back's time x and then say the way out's time was x+1.

It's important to know that R times T = D

Therefore we can say 300(x+1)=D and 350(x)=D

Since the distance is the same, 300(x+1)=350(x)

Solve:

300(x+1)=350x

300x+300=350x

50x=300

x=6

Therefore, the trip out took 6 hours.

I hope that helps! :)

Answer:the trip out took 7 hours

Step-by-step explanation:

Distance travelled = speed × time

Let t represent the time it took the plane to travel from California.

It took one hour on the way out than it did on the way back. This means that the time it took the plane to return would be t - 1

The plane's average speed out ward 300 mph. Distance travelled when going out would be

300 × t = 300 = 300t

The speed on the way back was 350mph. Distance travelled when returning would be

350(t - 1) = 350t - 350

Since the distance is the same, it means that

300t = 350t - 350

350t - 300t = 350

50t = 350

t = 350/50 = 7

Time taken to go out is 7 hours

Answer:

a

Step-by-step explanation:

i think its most suitable

Answer:

1020–1029 kg/m3

Step-by-step explanation:

The density of seawater at the surface of the ocean varies from 1,020 to 1,029 kilograms per cubic meter.

The density of seawater is a function of temperature, salinity, and pressure.

To measure the density of ocean water a sample of sea water is collected and brought into the laboratory to be measured. Salinity, temperature and pressure are measured to find density. The formula for measuring density is p(density) = m(mass) / V(volume).

Answer:

1020 to 1029.

Step-by-step explanation:

I searched it like it says in the unit actinity course Geometry semester B.

We can be 95% confident that the true proportion of all voters in the state who favor approval lies between 0.438 < p < 0.505. Option B is the correct answer.

Here's a breakdown of the steps involved in constructing the confidence interval:

Calculate the sample proportion (P):

P = 408 (voters favoring approval) / 865 (total voters surveyed) = 0.4714

Determine the critical value (z):

For a 95% confidence interval, the critical value from the standard normal distribution is 1.96.

Calculate the margin of error:

Margin of error = z * √(P * (1 - P) / n)

Margin of error = 1.96 * √(0.4714 * (1 - 0.4714) / 865) = 0.0335

Construct the confidence interval:

Lower bound = P - margin of error = 0.4714 - 0.0335 = 0.4379

Upper bound = P + margin of error = 0.4714 + 0.0335 = 0.5049

Round the bounds to three decimal places:

0.438 < p < 0.505

Therefore, we can be 95% confident that the true proportion of all voters in the state who favor approval lies between 0.438 and 0.505.

Answer:

b. [tex]( N^c \cap D ) \cup ( N \cap D^c )[/tex]

Step-by-step explanation:

Here,

N = { x ∈ U | x subscribes to News Magazine },

D = { x ∈ U | x subscribes to The Daily Informer },

Such that N∩D ≠ ∅,

Now, the number of the set of people surveyed who subscribe only News Magazine

= Subscribe News Magazine and not Daily Informer

= [tex]N\cap D^c[/tex]

Similarly, the number of the set of people surveyed who subscribe only Daily informer

= Subscribe Daily informer and not subscribe News Magazine

= [tex]D\cap N^c[/tex]

Hence,  the set of people surveyed who subscribe to exactly one of the two news sources given

= Subscribe News Magazine and not subscribe Daily Informer or subscribe Daily informer and not subscribe News Magazine

[tex]= (N\cap D^c)\cup (D\cap N^c)[/tex]

[tex]=( N^c \cap D ) \cup ( N \cap D^c )[/tex]

The set of people who subscribe to exactly one of the two news sources is represented by ( Nc ∩ D ) ∪ ( N ∩ Dc ). This includes people who subscribe to The Daily Informer but not News Magazine and those who subscribe to News Magazine but not The Daily Informer.

The set that represents the set of people surveyed who subscribe to exactly one of the two news sources, The Daily Informer (D) or News Magazine (N), can be written as ( Nc ∩ D ) ∪ ( N ∩ Dc ). The notation Nc represents the set of all people who do not subscribe to News Magazine and Dc represents the set of all people who do not subscribe to The Daily Informer. Thus, Nc ∩ D represents those who subscribe to The Daily Informer but not News Magazine, while N ∩ Dc represents those who subscribe to News Magazine but not The Daily Informer. The union of these two sets, denoted by ∪, represents the complete set of people who subscribe to exactly one of the two news sources.

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Answer:

The different single-scoop ice-cream cones can you buy from this vendor=12

Step-by-step explanation:

Given that a  vendor sells ice cream from a cart on the boardwalk. He offers vanilla, chocolate, strawberry, blueberry, and pistachio ice cream, served on either a waffle, sugar, or plain cone.

Different types of icecreams are 4

Types of cones are 3

For each ice cream say we can have either waffle, sugar or plain cone.

i.e. for each type of ice cream, we have 3 different cones.

Hence for 4 types of ice creams we have 3*4=12 different single scoop ice creamcones.

There are 15 different single-scoop ice-cream cones that can be bought from the vendor.

To calculate the number of different single-scoop ice-cream cones that can be bought from the vendor, we need to multiply the number of ice cream flavors with the number of cone types. The vendor offers 5 flavors and 3 cone types, so the total number of different single-scoop ice-cream cones that can be bought is 5 x 3 = 15.

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Answer:

Consider the following calculations

Step-by-step explanation:

Since 1 Blimp uses 2 components of B and C each

=> choosing 2 components of B(remaining after using in other prototypes) for 1st model= 22C2

choosing 2 components of B(remaining after using in other prototypes) for 2nd model= 21C2

choosing 2 components of B(remaining after using in other prototypes) for 3rd model= 20C2

choosing 2 components of B(remaining after using in other prototypes) for 4th model= 19C2

choosing 2 components of B(remaining after using in other prototypes) for 5th model= 18C2

and choosing 2 components of C(remaining after using in other prototypes) = 24C2

Similarly for C

P(5 prototypes of Blimp created)=[(22C2 / 25C2 )*(24C2 / 25C2 )] + [(21C2 / 25C2 )*(23C2 / 25C2 )]+[(20C2 / 25C2 )*(22C2 / 25C2 )]+[(19C2 / 25C2 )*(21C2 / 25C2 )]+[(18C2 / 25C2 )*(20C2 / 25C2 )]

Answer:

The cost of producing those 50 blu-rays is given by, 1337.5 unit and it is 26.75 unit for each blu-ray.

Step-by-step explanation:

The cost of producing x blu-rays is given by,

B(x) = 1250 + 1.75x unit -----------------------------(1)

So, cost of producing 50 blu-rays is given by,

B(50) = [tex]1250 + 50 \times (1.75)[/tex] unit

         = (1250 + 87.5) unit

         = 1337.5 unit

So, for each blu-ray, it is,

[tex]\frac {1337.5}{50}[/tex] unit

= 26.75 unit

Answer: Our required probability is 20%.

Step-by-step explanation:

Since we have given that

Probability of the employees needed corrective shoes = 8%

Probability of the employees needed major dental work = 15%

Probability of employees needed both corrective shoes and major dental work = 3%

So, we need to find the probability that an employee will need either corrective shoes or major dental work.

So, it becomes,

[tex]P(S\cup D)=P(S)+P(D)-P(S\cap D)\\\\P(S\cup D)=8+15-3\\\\P(S\cup D)=23-3\\\\P(S\cup D)=20\%[/tex]

Hence, our required probability is 20%.

Answer:

At least 81 observations should be made to be 95% confident that the observed data is within a tolerance of ±10 percent of the population data.

Step-by-step explanation:

The following equation is used to compute the minimum sample size required to estimate the population proportion  within the required margin of error:

n≥ p×(1-p) × [tex](\frac{z}{ME} )^2[/tex] where

Then, n≥ 0.30×0.70 × [tex](\frac{1.96}{0.10} )^2[/tex] ≈ 80.67

At least 81 observations should be made to be 95% confident that the observed data is within a tolerance of ±10 percent of the population data.

Random observations should include different employees, and sampling time should also be random.

Answer:

5

Step-by-step explanation:

Assuming the intelligence test had 10 questions, since Terry was average for his age group, his most likely IQ score is 5

Answer:

Not Independent , Independent , Not Independent

Step-by-step explanation:

Two events A and B are said to be independent when:

P(A ∩ B) = P(A) × P(B)

In the question P(A|B) is given i.e the probability of event A given that event B has already occured.

P(A|B) = P(A ∩ B) ÷ P(B)

If A and B are independent , the probability of A is not affected by B.

In terms of equation:

P(A|B) = P(A ∩ B)÷ P(B)

          = [P(A)×P(B)] ÷ P(B)

          = P(A)

Only in the second option we can see :

P(A|B) = P(A) = 0.2

Hence the events A and B are independent only in second case.

Answer:

a) The 90% confidence interval would be given by (230.212;233.128)

b) For this case if we analyze the confidence interval we see that not contains the value of 235. So we can't support the claim that the true mean is higher than 235 Hz at 10% of significance.

Step-by-step explanation:

Previous concepts  and data given

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".

Data: 230.66, 233.05, 232.58, 229.48, 232.58

We can calculate the mean and the deviation from these data with the following formulas:

[tex]\bar X= \frac{\sum_{i=1}^n x_i}{n}[/tex]

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}[/tex]

[tex]\bar X=231.67[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s=1.531 represent the sample standard deviation

n=5 represent the sample size  

Part a) Confidence interval

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=5-1=4[/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: "=-T.INV(0.05,4)".And we see that [tex]t_{\alpha/2}=2.13[/tex]

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

[tex]231.67-2.13\frac{1.531}{\sqrt{5}}=230.212[/tex]    

[tex]231.67+2.13\frac{1.531}{\sqrt{5}}=233.128[/tex]

So on this case the 90% confidence interval would be given by (230.212;233.128)

Part b)  Do the results of your calculations support the claim that mean natural frequency is 235 Hz?  

For this case if we analyze the confidence interval we see that not contains the value of 235. So we can't support the claim that the true mean is higher than 235 Hz at 10% of significance.

To determine the 90% confidence interval for mean natural frequency, calculate the mean, standard deviation, find the t-value, and compute the margin of error and CI. The claim that the mean frequency is 235 Hz is assessed by checking if this value is within the calculated CI.

To find the 90% two-sided confidence interval (CI) on mean natural frequency for the five delaminated beams, we first calculate the mean (\( \bar{x} \)) and the standard deviation (s) of the given frequencies. Then, we use the t-distribution, as the sample size is small (n=5), to find the t-value for a 90% CI, which corresponds to a significance level (alpha) of 0.10 and degrees of freedom (df) of n-1. Finally, we apply the formula for the 90% CI as follows:

(b) To assess whether the mean natural frequency is 235 Hz, we would check if this value lies within the 90% CI. If it does not, the results do not support the claim that the mean natural frequency is 235 Hz.

Answer:

Step-by-step explanation:

Given that researchers are studying the relationship between honesty, age, and self-control conducted an experiment on 160 children between the ages of 5 and 15.

a) The population is the  children between the ages of 5 and 15 in general

D. All children between the ages of 5 and 15

b) Sample is

D. 160 children between the ages of 5 and 15

c) Yes becauB. If the sample is randomly selected and representative of the entire population, then the results can be generalized to the target population. Furthermore, since this study is experimental, the findings can be used to infer causal relationships.se randomly selected and also sample size is above 30

The population of interest is all children aged 5 to 15; the study's sample consists of the 160 participating children. Results can be generalized if the sample is representative, and as an experimental study, it can establish causal relationships.

The population of interest in the study is D. All children between the ages of 5 and 15 because the researchers are attempting to understand a broader phenomenon that could be applicable to all children within this age range, not just those in the study. The sample for the study is D. 160 children between the ages of 5 and 15 because these are the actual participants that researchers have collected data from within this experiment.

Regarding the generalizability and establishment of causal relationships of the study, the correct option is B. If the sample is randomly selected and representative of the entire population, then the results can be generalized to the target population. Furthermore, since this study reports the manipulation of an independent variable (the instruction given to one group and not to the other), it qualifies as an experimental study, and its results can inform us about causal relationships.

Answer:  The required probabilities are

[tex](a)~0\\\\(b)\dfrac{1}{9},\\\\(c)~\dfrac{1}{36}.[/tex]

Step-by-step explanation:  Given that a pair of fair dice is rolled.

We are to find the probability of getting

(a) getting a sum of 1.

b) getting a sum of 5.

c) getting a sum of 12.

Let S be the sample space for the experiment of rolling a pair of fair dice.

Then, S = {(1,1), (1,2), (1,3), (1, 4), (1,5), (1,6), .  . . , (6,5), (6,6)}.

And, n(S) =36.

(a) Let E denote the event of getting a sum of 1.

Since the sum of the numbers on two dice is minimum 2, so

E = { }  ⇒  n(E) = 0.

Therefore, the probability of event E is

[tex]P(E)=\dfrac{n(E)}{n(S)}=\dfrac{0}{36}=0.[/tex]

(b) Let F denote the event of getting a sum of 5.

Then,

F = {(1,4), (2,3), (3,2), (4,1)}  ⇒  n(F) = 4.

Therefore, the probability of event F is

[tex]P(F)=\dfrac{n(F)}{n(S)}=\dfrac{4}{36}=\dfrac{1}{9}.[/tex]

(c) Let G denote the event of getting a sum of 12.

Then,

G = {(6,6)}  ⇒  n(G) = 1.

Therefore, the probability of event G is

[tex]P(G)=\dfrac{n(G)}{n(S)}=\dfrac{1}{36}.[/tex]

Thus, the required probabilities are

[tex](a)~0\\\\(b)\dfrac{1}{9},\\\\(c)~\dfrac{1}{36}.[/tex]

The probability of getting a sum of 1, the sum of 5, and the sum of 12 are 0, 1/9, and 1/36 respectively

Probability is the likelihood or chance that an event will occur.

For a pair of rolled dice, the total outcome will be 6² = 36

Probability = Expected outcome/Total outcome

a) Pr(getting a sum of 1) = 0/36 = 0

Note that since we have a pair of dice, the least sum we can have is 2

b) The event for getting a sum of 5 are (1, 4), (4, 1), (3, 2), (2, 3)

n(E) = 4

Pr( getting a sum of 5) = 4/36 = 1/9

c) The event for getting a sum of 12 are (6, 6)

n(E) = 1

Pr( getting a sum of 12) = 1/36

Hence the probability of getting a sum of 1, the sum of 5, and the sum of 12 are 0, 1/9, and 1/36 respectively

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Answer: the units digit has to be a 1 digit number not 2

Step-by-step explanation:

Answer:

Option e -  0.5597

Step-by-step explanation:

Given : Suppose that a random sample of size 49 is to be selected from a population with mean 49 and standard deviation 6.

To find :  What is the approximate probability that will be more than 0.5 away from the population mean?

Solution :  

Applying z-score formula,

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

Here, [tex]\mu=49,\sigma=6, n=49[/tex]

As, probability that will be more than 0.5 away from the population mean

i.e. x=49.5,

[tex]z=\frac{49.5-49}{\frac{6}{\sqrt{49}}}[/tex]

[tex]z=\frac{0.5}{0.857}[/tex]

[tex]z=0.58[/tex]

In z-score table the value of z at 0.58 is 0.7190.

x=48.5,

[tex]z=\frac{48.5-49}{\frac{6}{\sqrt{49}}}[/tex]

[tex]z=\frac{-0.5}{0.857}[/tex]

[tex]z=-0.58[/tex]

In z-score table the value of z at -0.58 is 0.2810.

Now, probability that will be more than 0.5 away from the population mean is given by,

[tex]P=P(X>49.5)+P(X<48.5)[/tex]

[tex]P=1-P(X<49.5)+P(X<48.5)[/tex]

[tex]P=1-0.7190+0.2810[/tex]

[tex]P=0.562[/tex]

Therefore, option e is correct.

Answer:

A.

[tex]H_0: \mu\leq514\\\\H_1: \mu>514[/tex]

B. Z=2.255. P=0.01207.

C. Although it is not big difference, it is an improvement that has evidence. The scores are expected to be higher in average than without the review course.

D. P(z>0.94)=0.1736

Step-by-step explanation:

A. state the null and alternative hypotheses.

The null hypothesis states that the review course has no effect, so the scores are still the same. The alternative hypothesis states that the review course increase the score.

[tex]H_0: \mu\leq514\\\\H_1: \mu>514[/tex]

B. test the hypothesis at the a=.10 level of confidence. is a mean math score of 520 significantly higher than 514? find the test statistic, find P-value. is the sample statistic significantly higher?

The test statistic Z can be calculated as

[tex]Z=\frac{M-\mu}{s/\sqrt{N}} =\frac{520-514}{119/\sqrt{2000}}=\frac{6}{2.661}=2.255[/tex]

The P-value of z=2.255 is P=0.01207.

The P-value is smaller than the significance level, so the effect is significant. The null hypothesis is rejected.

Then we can conclude that the score of 520 is significantly higher than 514, in this case, specially because the big sample size.

C.​ do you think that a mean math score of 520 vs 514 will affect the decision of a school admissions adminstrator? in other words does the increase in the score have any practical significance?

Although it is not big difference, it is an improvement that has evidence. The scores are expected to be higher in average than without the review course.

D. test the hypothesis at the a=.10 level of confidence with n=350 students. assume that the sample mean is still 520 and the sample standard deviation is still 119. is a sample mean of 520 significantly more than 514? find test statistic, find p value, is the sample mean statisically significantly higher? what do you conclude about the impact of large samples on the p-value?

In this case, the z-value is

[tex]Z=\frac{520-514}{s/\sqrt{n}} =\frac{6}{119/\sqrt{350}} =\frac{6}{6.36} =0.94\\\\P(z>0.94)=0.1736>\alpha[/tex]

In this case, the P-value is greater than the significance level, so there is no evidence that the review course is increasing the scores.

The sample size gives robustness to the results of the sample. A large sample is more representative of the population than a smaller sample. A little difference, as 520 from 514, but in a big sample leads to a more strong evidence of a change in the mean score.

Large samples give more extreme values of z, so P-values that are smaller and therefore tend to be smaller than the significance level.

A math teacher claims to have developed a review course that increases the score of students on the math portion of a college entrance exam. The null hypothesis is that there is no difference in scores before and after the course, while the alternative hypothesis is that the course increases the scores. The hypothesis is tested at the 0.10 level of confidence by calculating the z-score and finding the p-value. The impact of the score increase on the decision of a school admissions administrator depends on their criteria. The hypothesis is then tested again with a larger sample size, following the same steps as earlier.

A. The null hypothesis (H0) is that there is no difference in scores before and after the review course, while the alternative hypothesis (H1) is that the review course increases the scores on the math portion of the college entrance exam.

B. To test the hypothesis at the 0.10 level of confidence, we calculate the z-score and find the p-value. The test statistic is z = (sample mean - population mean) / (population standard deviation / sqrt(sample size)). The p-value is the probability of obtaining a test statistic as extreme as the observed value, assuming the null hypothesis is true. If the p-value is less than the significance level, we reject the null hypothesis and conclude that the mean math score is significantly higher than 514.

C. Whether a mean math score increase from 514 to 520 has any practical significance depends on the requirements set by the school admissions administrator. If the administrator determines that a small increase in score does not significantly impact their decision, then the increase may not have practical significance. However, if the administrator considers even small improvements in score as valuable, then the increase may have practical significance.

D. To test the hypothesis at the 0.10 level of confidence with a sample size of 350 students, we follow the same steps as in part B to calculate the test statistic and p-value. If the p-value is less than the significance level, we reject the null hypothesis and conclude that the sample mean of 520 is significantly higher than 514. Large sample sizes can result in smaller p-values, indicating stronger evidence against the null hypothesis.

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Answer:

a) Null hypothesis:[tex]\mu \leq 30.2[/tex]  

Alternative hypothesis:[tex]\mu > 30.2[/tex]  

b) [tex]X \sim N(\mu=32.12, \sigma=4.83)[/tex]

And the distribution for the random sample is given by:

[tex]\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)[/tex]

c) [tex]t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81[/tex]    

[tex]p_v =P(t_{(49)}>2.81)=0.0035[/tex]  

d) If we compare the p value and the significance level assumed [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.  

e) The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation  

[tex]\bar X=32.12[/tex] represent the sample mean  

[tex]s=4.83[/tex] represent the sample standard deviation

[tex]n=50[/tex] sample size  

[tex]\mu_o =30.2[/tex] represent the value that we want to test

[tex]\alpha[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

a. Define the parameter and state the hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 30.2 mpg, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 30.2[/tex]  

Alternative hypothesis:[tex]\mu > 30.2[/tex]  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

b. Define the sampling distribution (mean and standard deviation).

Let X the random variable who represent the variable of interest. And we know that the distribution for X is:

[tex]X \sim N(\mu=32.12, \sigma=4.83)[/tex]

And the distribution for the random sample is given by:

[tex]\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)[/tex]

c. Perform the test and calculate P-value

Calculate the statistic

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

[tex]t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81[/tex]    

P-value

The first step is calculate the degrees of freedom, on this case:  

[tex]df=n-1=50-1=49[/tex]  

Since is a one right tailed test the p value would be:  

[tex]p_v =P(t_{(49)}>2.81)=0.0035[/tex]  

d. State your conclusion.

If we compare the p value and the significance level assumed [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.  

e. Explain what the p-value means in this context.

The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.

Final answer:

Explanation of hypothesis testing for a fuel economy standard compliance claim.

Explanation:

Hypotheses:

Sampling Distribution:

Performing the Test:

Calculate the Z-score and P-value from the sample data using the formula. Compare the P-value to the significance level.

Conclusion: Since **p-value = 0.2578** is greater than the significance level of 0.05, we fail to reject the null hypothesis. There is not enough evidence to claim that the fleet meets the fuel economy standard.

Answer:

[tex](x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}[/tex]

Step-by-step explanation:

Given:

Center of circle is at (5, -4).

A point on the circle is [tex](x_1,y_1)=(-3, 2)[/tex]

Equation of a circle with center [tex](h,k)[/tex] and radius 'r' is given as:

[tex](x-h)^2+(y-k)^2=r^2[/tex]

Here, [tex](h,k)=(5,-4)[/tex]

Radius of a circle is equal to the distance of point on the circle from the center of the circle and is given using the distance formula for square of the distance as:

[tex]r^2=(h-x_1)^2+(k-y_1)^2[/tex]

Using distance formula for the points (5, -4) and (-3, 2), we get

[tex]r^2=(5-(-3))^2+(-4-2)^2\\r^2=(5+3)^2+(-6)^2\\r^2=8^2+6^2\\r^2=64+36=100[/tex]

Therefore, the equation of the circle is:

[tex](x-5)^2+(y-(-4))^2=100\\(x-5)^2+(y+4)^2=100[/tex]

Now, rewriting it in the form asked in the question, we get

[tex](x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}[/tex]

Answer:

94% Confidence interval:  (0.3771 ,0.5365)

Step-by-step explanation:

We are given the following data set:

0.481, 0.439,0.448,0.446,0.618, 0.411, 0.250, 0.604, 0.414

a) True mean of these distance estimates by d

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{4.111}{9} = 0.4568[/tex]

b)

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

Sum of squares of differences = 0.0954

[tex]S.D = \sqrt{\frac{0.0954}{8}} = 0.1092[/tex]

94% Confidence interval:  

[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]  

Putting the values, we get,  

[tex]t_{critical}\text{ at degree of freedom 8 and}~\alpha_{0.06} = \pm 2.1891[/tex]  

[tex]0.4568 \pm 2.1891(\frac{0.1092}{\sqrt{9}} ) = 0.4568 \pm 0.0797 = (0.3771 ,0.5365)[/tex]  

Final answer:

To calculate the 94% confidence interval for the true mean distance of an asteroid, one needs to compute the sample mean, standard error, and then apply the appropriate Z-score to determine the range.

Explanation:

The student's question is related to calculating a confidence interval for the true mean distance of a near-Earth asteroid. With the given unbiased estimates of the distances and a standard deviation of 15, we will calculate the 94% confidence interval for the true distance d.

Firstly, calculate the sample mean X by adding up all the given distance estimates and dividing by the number of observatories, which is 9.

Next, find the standard error (SE) of the mean by dividing the standard deviation by the square root of the number of observatories (n).

To determine the 94% confidence interval, we'll use the Z-score associated with it (approximately 1.88) and the standard error.

The confidence interval formula is X ± (Z * SE). We calculate the upper and lower bounds of the interval accordingly.

By following these steps, the student can establish the 94% confidence interval, giving them a range within which the true mean distance of the asteroid is likely to fall.

Answer:

0.6708 or 67.08%

Step-by-step explanation:

Helen can only make both free throws if she makes the first. The probability that she makes the first free throw is P(C) = 0.78, now given that she has already made the first one, the probability that she makes the second is P(D|C) = 0.86. Therefore, the probability of Helen making both free throws is:

[tex]P(C+D) =  P(C) *P(D|C) = 0.78*0.86\\P(C+D) = 0.6708[/tex]

There is a 0.6708 probability that Helen makes both free throws.

Answer:

Option A is right

Step-by-step explanation:

Given that an  article in an educational journal discussing a non-conventional high school reading program reports a 95% confidence interval of (520 , 560) for the average score on the reading portion of the SAT.

Let X be the  score on the reading portion of the SAT.

By central limit theorem x bar, the sample mean will follow a normal.

95% confidence interval shows that mean

= average of the two limits

= 540

Margin of error = upper bound of confidence interval - Mean

=[tex]560-540=20[/tex]

Option A is right

That figure obviously doesn't go with this problem.  It doesn't matter; this is triangle ABC labeled the usual way.

c = 10, B = 35°, C = 65°

We have two angles and a side.  The third angle is obviously

A = 180° -  35°- 65°  = 80°

The remaining sides are given by the Law of Sines,

[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]

[tex]a = \dfrac{c \sin A}{\sin C} = \dfrac{10 \sin 80}{\sin 65} = 10.866[/tex]

[tex]b = \dfrac{c \sin B}{\sin C} = \dfrac{10 \sin 35}{\sin 65} = 6.328[/tex]

Answer: A=80°, a=10.9, b=6.3, third choice

Final answer:

A line parallel to one with a slope of -5/8 has the same slope, -5/8, while the slope of a line perpendicular to it is the negative reciprocal, which is 8/5.

Explanation:

The slope of a line that is parallel to a given line is identical to that of the given line. Therefore, a line parallel to one with a slope of -5/8 would also have a slope of -5/8. However, a line that is perpendicular to the given line would have a slope that is the negative reciprocal of the original slope.

To find the negative reciprocal, you flip the fraction and change the sign. So, the slope of a line perpendicular to a line with a slope of -5/8 would be 8/5.

[tex]

\begin{cases}

f(x)=x/3+7\\

f(x)=13

\end{cases}

[/tex]

We find that

[tex]13=x/3+7\Longrightarrow x=\boxed{18}[/tex]

Hope this helps.