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3 years ago

Solve. Iready pls help!!!

Mathematics

1 answer:

andrew11 [14]3 years ago

8 0

Answer:

Step-by-step explanation:

Also so you never gotta ask again use a website called wolframalpha it helped me

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PLEASE HELP!!!!!<br><br> find the area of the polygon.

CaHeK987 [17]

Answer:

2 cm is were youll find the polygon at. it shouldnt be hard to see the area of were the polygon is

7 0

2 years ago

Read 2 more answers

You are curious about the average number of yards Matthew Stafford throws for each game for the Detroit Lions. You randomly sele

Damm [24]

Answer:

The margin of error for this estimate is of 14.79 yards per game.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

T interval

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 20 - 1 = 19

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 2.093

The margin of error is:

$M = T\frac{s}{\sqrt{n}}$

In which s is the standard deviation of the sample and n is the size of the sample.

You randomly select 20 games and see that the average yards per game is 273.7 with a standard deviation of 31.64 yards.

This means that $n = 20, s = 31.61$

What is the margin of error for this estimate?

$M = T\frac{s}{\sqrt{n}}$

$M = 2.093\frac{31.61}{\sqrt{20}}$

$M = 14.79$

The margin of error for this estimate is of 14.79 yards per game.

3 0

3 years ago

The mean SAT score in mathematics, M, is 600. The standard deviation of these scores is 48. A special preparation course claims

kupik [55]

Answer:

Step-by-step explanation:

The mean SAT score is $\mu=600$ , we are going to call it \mu since it's the "true" mean

The standard deviation (we are going to call it $\sigma$ ) is

$\sigma=48$

Next they draw a random sample of n=70 students, and they got a mean score (denoted by $\bar x$ ) of $\bar x=613$

The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.

- So the Null Hypothesis $H_0:\bar x \geq \mu$

- The alternative would be then the opposite $H_0:\bar x < \mu$

The test statistic for this type of test takes the form

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}$

and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.

With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}\\\\= \frac{| 600-613 |}{48/\sqrt(70}}\\\\= \frac{| 13 |}{48/8.367}\\\\= \frac{| 13 |}{5.737}\\\\=2.266\\$

<h3>since 2.266>1.645 we can reject the null hypothesis.</h3>