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Slav-nsk [51]
4 years ago
14

If 180°<α<270°, cos ⁡α= −8/17, 270°<β<360°, and sin β= −4/5, what is sin⁡(α+β)?

Mathematics
1 answer:
mars1129 [50]4 years ago
7 0

Answer:

\sin(\alpha+\beta) = -0.153

Step-by-step explanation:

Let determine the angles behind each trigonometric expression:

\cos \alpha = -\frac{8}{17}

\alpha = \cos^{-1}\left(-\frac{8}{17} \right)

Given that 180^{\circ}< \alpha < 270^{\circ}, the value of \alpha is:

\alpha \approx 241.928^{\circ}

\sin \beta = -\frac{4}{5}

\beta = \sin^{-1}\left(-\frac{4}{5} \right)

Given that 270^{\circ}< \beta, the value of \beta is:

\beta \approx 306.870^{\circ}

The sine function of the sum of angles can be determined by the following identity:

\sin(\alpha + \beta)=\sin \alpha \cdot \cos \beta + \sin \beta \cdot \cos \alpha

If \alpha \approx 241.928^{\circ} and \beta \approx 306.870^{\circ}, then:

\sin (241.928^{\circ}+306.870^{\circ}) = (\sin 241.928^{\circ}) \cdot (\cos 306.870^{\circ}) + (\sin 306.870^{\circ})\cdot (\cos 241.928^{\circ})\sin(241.928^{\circ}+306.870^{\circ}) = -0.153

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Step-by-step explanation:

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5(x² + 6x + 10)

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In a survey of a group of​ men, the heights in the​ 20-29 age group were normally​ distributed, with a mean of 67 inches and a s
liq [111]

Answer:

a) 20.33% probability that the participant is less than 64.5 inches.

b) 33.72% probability that the participant is more than 68.25 inches

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 67, \sigma = 3

a.) Find the probability that the participant is less than 64.5 inches?

This is the pvalue of Z when X = 64.5.

Z = \frac{X - \mu}{\sigma}

Z = \frac{64.5 - 67}{3}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033

20.33% probability that the participant is less than 64.5 inches.

b.) Find the probability that the participant is more than 68.25 inches?

This is 1 subtracted by the pvalue of Z when X = 68.25.

Z = \frac{X - \mu}{\sigma}

Z = \frac{68.25 - 67}{3}

Z = 0.42

Z = 0.42 has a pvalue of 0.6628

1 - 0.6628 = 0.3372

33.72% probability that the participant is more than 68.25 inches

3 0
4 years ago
Solve the equation (0°≤A≤360°)<br>a) √3sinA-cosA=√2​
Blizzard [7]

Step-by-step explanation:

√3sinA-cosA=√2

dividing both side by (√3²-1²=2 we get)

√3/2 sinA-1/2.cos A=√2/2

cos 30sinA-sin30cosA=√2/(√2×√2)

sinAcos30-cosAsin30=1/√2

sin(A-30)=sin45,sin(180-45)

therefore

A-30=45,135

A=45+30=75 or A=135+30=165

6 0
3 years ago
Amy has a triangle with side lengths of 6 and 8. What is the length of the <br> hypotenuse.
Alex787 [66]

Hello There!

The length of the hypotenuse is 10.

We know what 2 sides are 6 and 8.

To find the third side we use the formula a^2 + b^2 = c^2

So we multiply 6 by 6 which equals 36 and then multiply 8 by 8 which equals 64.

Next we add 64 and 36 together and we get 100.

Next, we square root 100 and get 10.

10 is the length of the hypotenuse.

8 0
3 years ago
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