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Alex73 [517]
2 years ago
10

In ΔHIJ, the measure of ∠J=90°, JI = 5, IH = 13, and HJ = 12. What is the value of the sine of ∠H to the nearest hundredth?

Mathematics
2 answers:
Arte-miy333 [17]2 years ago
3 0

<u>Given</u>:

Given that HIJ is a right triangle.

The measure of ∠J is 90°, JI = 5, IH = 13, and HJ = 12.

We need to determine the value of sine of ∠H

<u>Value of sine ∠H:</u>

The value of sine ∠H can be determined by using the trigonometric ratios.

Thus, we have;

sin \ H=\frac{JI}{IH}

Substituting the values, we get;

sin \ H=\frac{5}{13}

Dividing, we get;

sin \ H=0.3846

Taking sin^{-1} on both sides, we have;

H=sin^{-1}(0.3846)

H=22.6189

Rounding off to the nearest hundredth, we get;

H=22.62^{\circ}

Thus, the measure of ∠H is 22.62°

lesya692 [45]2 years ago
3 0

Answer:

.38

Step-by-step explanation:

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Answer:

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The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

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For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

Step-by-step explanation:

Information given

data: 202.2 203.4 200.5 202.5 206.3 198.0 203.7 200.8 201.3 199.0

We can calculate the sample mean and deviation with the following formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

\sigma=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=201.77 represent the sample mean    

s=2.41 represent the sample standard deviation    

n=10 sample size    

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\alpha=0.05 represent the significance level for the hypothesis test.    

t would represent the statistic

p_v represent the p value for the test

Hypothesis to test

We want to determine if the true mean is equal to 200, the system of hypothesis are :    

Null hypothesis:\mu = 200    

Alternative hypothesis:\mu = 200    

The statistic for this case is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)    

The statistic is given by:

t=\frac{201.77-200}{\frac{2.41}{\sqrt{10}}}=2.32    

The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

p_v =2*P(t_{(9)}>2.32)=0.0455    

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

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that means step 1 is strange... it didn't distribute properly

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