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3 years ago

[[ 20 POINTS IF YOU ANSWER THIS QUESTION CORRECTLY. ]]

Mathematics

1 answer:

dezoksy [38]3 years ago

3 0

Answer: the height of the kite is 106.065 ft

Step-by-step explanation:

The length of the kite represents the hypotenuse of the right angle triangle. The height of the kite represents the opposite side of the right angle triangle.

To determine x, the height of the kite, we would apply the sine trigonometric ratio which is expressed as

Sin θ = opposite side/hypotenuse

Therefore,

Sin 45 = x/150

Cross multiplying, it becomes

x = 150Sin45 = 150 × 0.7071

x = 106.065 ft

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Jack and Jill fetched 24 pints of water in their bucket. Jill drank a tenth of this, jack a quarter. They gave three eighths to

Minchanka [31]

Answer: See explanation

Step-by-step explanation:

Total water fetched = 24 pints.

Jill drank a tenth of this. This will be:

= 1/10 × 24

= 2.4 pints

Jack drank a quarter. This will be:

= 1/4 × 24 = 6.

They gave three eighths to dame Dob . Thus will be:

= 3/8 × 24

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They gave a fifth to jill's mother. This will be:

= 1/5 × 24.

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2 years ago

A gondola (cable car) at a ski area holds 50 people. Its maximum safe load is 10000 pounds. A population of skiers has a distrib

fenix001 [56]

Answer: 0.03855

Step-by-step explanation:

Given :A population of skiers has a distribution of weights with mean 190 pounds and standard deviation 40 pounds.

Its maximum safe load is 10000 pounds.

Let X denotes the weight of 50 people.

As per given ,

Population mean weight of 50 people = $\mu=50\times190=9500\text{ pounds}$

Standard deviation of 50 people $=\sigma=40\sqrt{50}=40(7.07106781187)=282.84$

Then , the probability its maximum safe load will be exceeded =

$P(X>10000)=P(\dfrac{X-\mu}{\sigma}>\dfrac{10000-9500}{282.84})\\\\=P(z>1.7671-8)\\\\=1-P(z\leq1.7678)\ \ \ \ [\because\ P(Z>z)=P(Z\leq z)]\\\\=1-0.96145\ \ \ [\text{ By p-value of table}]\\\\=0.03855$

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3 years ago

Does this set of ordered pairs represent a function? (-1,5)(0,-3)(2,7)(4,0)(7,5)

ruslelena [56]

Answer:

Step-by-step explanation:

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2 years ago

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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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3 years ago