All categories

3 years ago

Cos y = 8/16 and round your answer to the nearest tenth

Mathematics

1 answer:

juin [17]3 years ago

6 0

Answer:

Step-by-step explanation:

To find y, notice it is equal to 8 / 16 or 1/2. This is a value from a special triangle 30,60, 90. This means y = 60. 1 is always opposite 30 and this equation needs 1 to be adjacent.

You might be interested in

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

Please find the answer

jek_recluse [69]

Answer:

I hope this is correct but 8.5 or 8 1/2 Units

5 0

3 years ago

Read 2 more answers

How can i find the área of a triangle? please help.

Anni [7]

This is the area of a triangle

7 0

3 years ago

Read 2 more answers

At the gift shop, they sell small greeting cards and large greeting cards. The cost of a small greeting card is $1.60 and the co

erik [133]

<h2>Explanations:</h2>

From the given question, we have the following parameters

Cost of a small greeting card = $1.60

The cost of a large greeting card = $4.05.

Cost of 5 small greeting cards = 5(1.60)

Cost of 5 small greeting cards = $8.00

Cost of 4 large greeting cards = 4(4.05)

Cost of 4 large greeting cards = $16.2

Cost of 5 small and 4 large = $8.00 + $16.2

Cost of 5 small and 4 large $24.2

Hence it would cost $24.2 to get 5 small greeting cards and 4 large greeting cards

Similarly;

Cost of x small greeting cards = 1.60x

Cost of y large greeting cards = 4.05y

Cost of small greeting cards and y large greeting cards = 1.60x + 4.05y

Hence it would cost $(1.60x + 4.05y) to get x small greeting cards and y large greeting cards

5 0

1 year ago

A telecommunication company has 7 satellites, two of which are sending a week signal. If two are picked at random without replac

Setler [38]

2/7 is the probability that the first satellite sends a weak signal. That leaves 6, one of which sends a weak signal. So the probability that the second also sends a weak signal is 1/6. Combine these and we get 2/7×1/6=1/21.

5 0

3 years ago