Y=x+5<br> x-5y=-9<br> Can someone please help ASAP

Anybody know the answers? A-? B-4 c-12 1/3?

shutvik [7]

Answer:

11.666666-repeated on and on

Step-by-step explanation:

but you can probably round to 11.7

7 0

3 years ago

Last week, Gina's art teacher mixed 9 pints of red paint with 6 pints of white paint to make pink. Gina mixed 4 pints of red pai

cricket20 [7]

No, in order to do the correct ratio she would have needed to add 4.5 pints to 3 pints because if you divide 9:6 by 2 it is 4.5:3 not 4:3

6 0

3 years ago

Solve for b: 2(b + 3) = b – 4

Reil [10]

Answer:

b = -10

Step-by-step explanation:

2(b + 3) = b – 4

2b + 6 = b - 4

-b -b

b + 6 = -4

-6 -6

b = -10

8 0

3 years ago

Read 2 more answers

Paul jogged 3.25 miles, rick jogged 3 1/3 miles and sean jogged 3 1/8 miles. list the boys from least to greatest distance jogge

Novosadov [1.4K]

Answer:

Sean jogged 3.125, Paul jogged a distance of 3.25, Rick jogged 3.33

Step-by-step explanation:

make the fractions into decimals once they are in decimal form you cna then go and place them into the correct order.

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

Y=x+5 x-5y=-9 Can someone please help ASAP