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2 years ago

Drag each statement to show whether it is true, based on the graph.

Mathematics

1 answer:

Elena-2011 [213]2 years ago

8 0

The statements that are true about this graph are the pumpkin grew at a rate of 6 ounces per week, and in 3 weeks the pumpkin grew to a total of 18 ounces.

The number 3 indicates the value in x, while the number 18 indicates the value of y.

x = 3 weeks (time)
y = 18 ounces (weight)

<h3>What statements are true about this graph?</h3>

It is true the rate was 6 ounces per week (18 ounces/6 = 3).
It is true by the end of week 3 the weight was 18 ounces.

Learn more about graphs in: brainly.com/question/16608196

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A dinner bill came to $38.75 What is the total bill after 6% sales tax and 20% tip?

Natasha2012 [34]

38.75*6%=2.33
38.75*20%=7.75
7.75+2.33+38.75=48.83
I did tax and tip separately because you don't add tax or tip together like (38.75+2.33)*20% because that be paying more for nothing.

4 0

4 years ago

If a sprinkler waters 1/12 of a lawn in 1/4 hour, how much time will it take to water the entire lawn?

earnstyle [38]

(1/4 hour)/(1/12 lawn) = 3 hour/lawn

Selection D is appropriate.

7 0

4 years ago

Read 2 more answers

Solve for x. 12x = 2x^2 + 18

Zanzabum

2x^2-12x+18=0

Factor out GCF
2(x^2-6x+9)

Factor trinomial
(x-3)(x-3)

Proof
-3-3=-6
-3^2=9

Zeros
x-3=0
x=3

Final answer: B

7 0

4 years ago

Select the correct answer.

Galina-37 [17]

Answer:

The answer is C, hope that helps:)

Step-by-step explanation:

5 0

4 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

4 years ago