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3 years ago

Find sin θ Right Triangle Trigonometry

Mathematics

2 answers:

arlik [135]3 years ago

5 0

Answer:

Step-by-step explanation:

Use SOHCAHTOA

We are given the opposite side and the agacent side

Therefore we use TOA

tan(x)= 15/8

tan^-1(15/8)= 61.93

sin(61.93)= 15/17

Hope this Helped :)

creativ13 [48]3 years ago

3 0

Answer:

15/17

Step-by-step explanation:

15^2+8^2=c^2

c=hypotenuse. Sin is opposite side/hypotenuse

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HELP PLEASEEEEEEEEEEEEEE

vazorg [7]

Answer:

-2/4, 8/40, 21/30, 9, 9 20/25

Step-by-step explanation:

9/20/25 is a bit greater than 9.

21/30 is 9 away from a whole 8/40 is 32 away from 40 -2/4 is way less because its on the negative side which is like so far away from the other numbers.

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3 years ago

PLEASE HELLP ME! I would appreciate if you can!!

S_A_V [24]

Answer:

(2) A

Step-by-step explanation: easy Math

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3 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

7 0

3 years ago

A poll found that 53% of a sample of 1170 teens in a certain large country go online several times a day. (Treat this sample as

posledela

Answer:(a) margin error = 2.4%

(b) The margin error gives the measure in percentage of how the population parameter determined differ from the real population statistics or value.

(c) in 90% of the samples of teens in the country, the percent who go online several times a day will be within 50.6% and 55.4%. of the estimated 100%

Step-by-step explanation:

Using the proportion formulae

Margin error = z √p(1-p)/n

n= 1170, p = 53% = 0.53, 1-p = 0.47

and the z value at 90% C.I = 1.645

M error= 1.645 √0.53×0.47/1170

Margin error = 0.024 = 0.024 ×100

Margin error = 2.4%

53 - 2.4 = 50.6% and 53 + 2.4 = 55.4%

In other words 90% of the time: the number of teens who go online several time a day will be between 50.6 and 55.4%.

8 0

3 years ago

The first day of Sam's vacation was a Monday.

hichkok12 [17]

b is your answer today because it is the right anawer

5 0

3 years ago

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