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3 years ago

The space shuttle travels at 28000 mph while orbiting the earth. How far does the shuttle travel in feet per day

Mathematics

1 answer:

STatiana [176]3 years ago

6 0

We are given the speed of the shuttle as:

speed = 28,000 miles / hour

We know that:

1 mile = 5280 ft

24 hour = 1 day

Therefore by conversion:

speed = (28,000 miles / hr) * (5280 ft / 1 mile) * (24 hr / day)

speed = 3,548,160,000 ft / day

In 2 significant figures:

<span>speed = 3.5 x 10^9 ft / day</span>

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What is the length of the hypotenuse?

Rainbow [258]

Answer:

Hello! answer: 58

Step-by-step explanation:

40 × 40 = 1600

42 × 42 = 1764

1600 + 1764 = 3364

√3364 = 58

58 × 58 = 3364 therefore c = 58 Hope that helps!

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3 years ago

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(1+i)^2(x+yi)=4+5i<br> Solve for x and y

faust18 [17]

Answer:

x = 5/2

y = -2

Step-by-step explanation:

(1+i)² (x+yi) = 4+5i

(1+2i+i²)(x+yi) = 4+5i

(1+2i-1)(x+yi) = 4+5i

2i(x+yi) = 4+5i

2i²y + 2xi = 4+5i

-2y + 2xi = 4 + 5i

By comparing both sides,

-2y = 4

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x = 5/2

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2 years ago

The system of equations has 2x-y=2(x+1) y=-2

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Answer:

Step-by-step explanation:

This equation has no solution.

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2 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

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