How in the world. Im terrible at math

Using the correct graph, describe the association shown. Will mark brainliest❤️

olchik [2.2K]

Answer:

it is linear association

Step-by-step explanation:

3 0

3 years ago

A movie theater charges $5 for an adult’s ticket and $2 for a child’s ticket. one saturday, the theater sold 785 tickets for $32

laila [671]

They sold 570 adult tickets and 215 child tickets.

6 0

3 years ago

Read 2 more answers

One environmental group did a study of recycling habits in a California community. It found that 75% of the aluminum cans sold i

AysviL [449]

Answer:

a

$P( \^ p > 0.775 ) = 0.12798$

b

$P( 0.6718 < p < 0.775 ) =0.87183$

Step-by-step explanation:

From the question we are told that

The population proportion is $p = 0.75$

Considering question a

The sample size is $n = 387$

Generally the standard deviation of this sampling distribution is

$\sigma = \sqrt{ \frac{p(1 - p)}{ n } }$

=> $\sigma = \sqrt{ \frac{0.75(1 - 0.75)}{ 387 } }$

=> $\sigma = 0.022$

The sample proportion of cans that are recycled is

$\^ p = \frac{ 300}{387 }$

=> $\^ p = 0.775$

Generally the probability that 300 or more will be recycled is mathematically represented as

$P( \^ p > 0.775 ) = P( \frac{\^ p - p }{ \sigma } > \frac{0.775 - 0.75 }{ 0.022} )$

$\frac{\^ p - p }{\sigma } = Z (The \ standardized \ value\ of \ \^ p )$

$P( \^ p > 0.775 ) = P( Z > 1.136 )$

From the z table the area under the normal curve to the left corresponding to 1.591 is

$P( Z > 1.136) = 0.12798$

=> $P( \^ p > 0.775 ) = 0.12798$

Considering question b

Generally the lower limit of sample proportion of cans that are recycled is

$\^ p_1 = \frac{ 260 }{387 }$

=> $\^ p_1 = 0.6718$

Generally the upper limit of sample proportion of cans that are recycled is

$\^ p_2 = \frac{ 300}{387 }$

=> $\^ p_2 = 0.775$

Generally probability that between 260 and 300 will be recycled is mathematically represented as

$P( 0.6718 < p < 0.775 ) = P( \frac{0.6718 - 0.75 }{ 0.022}< \frac{\^ p - p }{ \sigma }$

=> $P( 0.6718 < p < 0.775 ) = P( -3.55 < Z < 1.136 )$

=> $P( 0.6718 < p < 0.775 ) = P(Z < 1.136 ) - P( Z < -3.55 )$

From the z table the area under the normal curve to the left corresponding to 1.136 and -3.55 is

$P( Z < -3.55 ) = 0.00019262$

and

$P(Z < 1.136 ) = 0.87202$

So

$P( 0.6718 < p < 0.775 ) = 0.87202- 0.00019262$

=> $P( 0.6718 < p < 0.775 ) =0.87183$

4 0

3 years ago

Several programs attempt to address the shortage of qualified teachers by placing uncertified instructors in schools with acute

ss7ja [257]

Answer:

We conclude that the mean scores with uncertified teachers is higher or equal as compared to certified teachers.

Step-by-step explanation:

We are given that reading scores of the students of certified teachers averaged 35.62 points with standard deviation 9.31. The scores of students instructed by uncertified teachers had mean 32.48 points with standard deviation 9.43 points on the same test.

There were 44 students in each group.

Let $\mu_1$ = mean scores with uncertified teachers.

$\mu_2$ = mean scores with certified teachers.

So, Null Hypothesis, $H_0$ : $\mu_1\geq \mu_2$ {means that the mean scores with uncertified teachers is higher or equal as compared to certified teachers}

Alternate Hypothesis, $H_A$ : $\mu_1$ {means that the mean scores with uncertified teachers is lower as compared to certified teachers}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean scores of students instructed by uncertified teachers = 32.48 points

$\bar X_2$ = sample mean scores of students instructed by certified teachers = 35.62 points

$s_1$ = sample standard deviation of scores of students instructed by uncertified teachers = 9.43 points

$s_2$ = sample standard deviation of scores of students instructed by certified teachers = 9.31 points

$n_1$ = sample of students under uncertified teachers = 44

$n_2$ = sample of students under certified teachers = 44

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(44-1)\times 9.43^{2} +(44-1)\times 9.31^{2} }{44+44-2} }$ = 9.37

So, the test statistics = $\frac{(32.48-35.62)-(0)}{9.37 \times \sqrt{\frac{1}{44} +\frac{1}{44} } }$ ~ $t_8_6$

= -1.572

The value of t test statistics is -1.572.

Since, in the question we are not given the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical values of -1.665 at 86 degree of freedom for left-tailed test.

Since our test statistic is more than the critical values of t as -1.572 > -1.665, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the mean scores with uncertified teachers is higher or equal as compared to certified teachers.

7 0

3 years ago

Please send answers!!

xz_007 [3.2K]

Answer:

Please refer to the above attachment...

(sorry for bad handwriting:P)...

Step-by-step explanation:

I hope it helps...:)

4 0

3 years ago