Liza is one year older than her sister.

Consider the bank account transactions shown. Which transaction will create the biggest change in the bank account balance?

arsen [322]

A withdrawal of $60

60 is the largest number therefore it’ll make the biggest impact because you’re loosing the most money

4 0

3 years ago

Which of the interpretations for the given expression is correct 5(3x-4)^2

Flauer [41]

<span>Which of the interpretations for the given expression is correct 5(3x-4)^2

is "5 times the square of the binomial 3x - 4."</span>

3 0

4 years ago

Read 2 more answers

Expand( a/3 - b/6 -1)^2

Artemon [7]

Answer:

Hello!

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( a/3 - b/6 -1)^2 as expanded form:

$\frac{a^2}{9} - \frac{ab}{9} - \frac{b^2}{36} - \frac{2a}{3} - \frac{b}{3} + 1$

Step-by-step explanation: You have to Simplify the expression.

Hope this helped you. Brainliest would be nice!

3 0

4 years ago

A newspaper advisor claims that an advertisement would increase sales by 35%. Sales are currently $500 per month. What will be t

dimulka [17.4K]

35% = 0.35
500 x 0.35 = 175
500 + 175 = 675 per month
675 x (12 x 3) = $24,300 sales over three years (I’m not sure what answer is wanted for this question)

3 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