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3 years ago

An athletic club charges a monthly membership

1 answer:

7 0

Answer: D

Explanation:

Special case:

New membership cost 65 but also free 2 class. However, the cost of a new member who takes x class where x > 2 which mean 2 or more class.

C(x) = 65$ + 15(x - 2)$

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Find an equation for the nth term of the arithmetic sequence. -15, -6, 3, 12, ...

Mariulka [41]

Answer: $t_{n}=-15+9(n-1)$

Step-by-step explanation:

The formula for finding the nth term of an arithmetic sequence is given as:

$t_{n}=a+(n-1)d$

$a =$ first term = -15

$d =$ common difference = -6 - (-15) = 9

$n =$ number of terms

substituting into the formula , we have :

$t_{n} = -15+(n-1)9$

$t_{n}=-15+9(n-1)$

5 0

3 years ago

Many elementary school students in a school district currently have ear infections. A random sample of children in two different

Marta_Voda [28]

Answer:

Step-by-step explanation:

The summary of the given data includes;

sample size for the first school $n_1$ = 42

sample size for the second school $n_2$ = 34

so 16 out of 42 i.e $x_1$ = 16 and 18 out of 34 i.e $x_2$ = 18 have ear infection.

the proportion of students with ear infection Is as follows:

$\hat p_1 = \dfrac{16}{42}$ = 0.38095

$\hat p_2 = \dfrac{18}{34}$ = 0.5294

Since this is a two tailed test , the null and the alternative hypothesis can be computed as :

$H_0 :p_1 -p_2 = 0 \\ \\ H_1 : p_1 - p_2 \neq 0$

level of significance ∝ = 0.05,

Using the table of standard normal distribution, the value of z that corresponds to the two-tailed probability 0.05 is 1.96. Thus, we will reject the null hypothesis if the value of the test statistics is less than -1.96 or more than 1.96.

The test statistics for the difference in proportion can be achieved by using a pooled sample proportion.

$\bar p = \dfrac{x_1 +x_2}{n_1 +n_2}$

$\bar p = \dfrac{16 +18}{42 +34}$

$\bar p = \dfrac{34}{76}$

$\bar p = 0.447368$

$\bar p + \bar q = 1 \\ \\ \bar q = 1 -\bar p \\ \\\bar q = 1 - 0.447368 \\ \\\bar q = 0.552632$

The pooled standard error can be computed by using the formula:

$S.E = \sqrt{ \dfrac{ \bar p \bar q}{ n_1} + \dfrac{\bar p \bar p}{n_2} }$

$S.E = \sqrt{ \dfrac{ 0.447368 * 0.552632}{ 42} + \dfrac{ 0.447368 * 0.447368}{34} }$

$S.E = \sqrt{ \dfrac{ 0.2472298726}{ 42} + \dfrac{ 0.2001381274}{34} }$

$S.E = \sqrt{ 0.01177284105}$

$S.E = 0.1085$

The test statistics is ;

$z = \dfrac{\hat p_1 - \hat p_2}{S.E}$

$z = \dfrac{0.38095- 0.5294}{0.1085}$

$z = \dfrac{-0.14845}{0.1085}$

z = - 1.368

Decision Rule: Since the test statistics is greater than the rejection region - 1.96 , we fail to reject the null hypothesis.

Conclusion: There is insufficient evidence to support the claim that a difference exists between the proportions of students who have ear infections at the two schools

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csqrt%7B0.81%7D" id="TexFormula1" title="\sqrt{0.81}" alt="\sqrt{0.81}" align="absmiddle" cl

Mandarinka [93]

Answer: .9

Step-by-step explanation:

3 0

3 years ago

Helpppp]ppppppppppppp

madam [21]

Answer:

I think it's b or d