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3 years ago

Find the measure of angle 4 AND...… angle 7

Mathematics

1 answer:

zaharov [31]3 years ago

4 0

Answer:

<4 = 126

<7 = 55

Step-by-step explanation:

Remark

There are only 2 answers to this question. <4 has one of them and <7 has the other one. <7 is acute. <2 is obtuse.

Givens

<2 = 55 degrees

Solution

<4 is the supplement of <2

Therefore <4 + <2 = 180 A straight line has an angle of 180°

<4 + 55 = 180

<4 = 180 - 55

<4 = 125

So <7 is either 55 or 125. Since <7 is acute, it is 55 degrees but we have another way of proving it.

<4 + <6 = 180 degrees Interior angles on the same side of the transversal are supplementary

125 + <6 = 180

<6 = 180 - 125

<6 = 55

<7 =<6 Vertically opposite angles are equal.

<7 = 55 degrees.

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Step-by-step explanation:

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Rewrite the quadratic function in intercept or factored form g(m)=3m^2+5-8m

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Split the second term in 3m^2 + 5 - 8m into two terms

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3 years ago

What is 5π/6 radians converted to degrees? If necessary, round your answer to the nearest degree.

klemol [59]

Answer:

300 degrees

Step-by-step explanation:

we know that $\pi = 360 degrees$ $\pi = 360 \ degrees$

we have to find value of $5\pi /6$

for that we have to multiply 5/6 to $\pi$ and 360 degrees

$\pi = 360 \ degrees\\5\pi /6 = 5/6 *360 = 300$

Thus, $5\pi /6$ = 300 degrees.

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3 years ago

Replacement times for TV sets are normally distributed with a mean of 8.2 years and a standard deviation of 1.1 years (Consumer

grandymaker [24]

Answer:

There is a 0.18% probability that a randomly selected TV will have a replacement time less than 5.0 years.

To provide a warranty so that only 1% of the TV sets will be replaced before the warranty expires, the length of the warranty is 10.76 years.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

Replacement times for TV sets are normally distributed with a mean of 8.2 years and a standard deviation of 1.1 years. This means that $\mu = 8.2, \sigma = 1.1$ .

Find the probability that a randomly selected TV will have a replacement time less than 5.0 years.

This is the pvalue of Z when $X = 5$ .

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

$Z = \frac{5 - 8.2}{1.1}$

$Z = -2.91$

$Z = -2.91$ has a pvalue of 0.00181

This means that there is a 0.18% probability that a randomly selected TV will have a replacement time less than 5.0 years.

If you want to provide a warranty so that only 1% of the TV sets will be replaced before the warranty expires, what is the time length of the warranty?

This is the value of X when Z has a pvalue of 0.99. This is $Z = 2.33$ .

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

$2.33 = \frac{X - 8.2}{1.1}$

$X - 8.2 = 2.33*1.1$

$X = 10.76$

To provide a warranty so that only 1% of the TV sets will be replaced before the warranty expires, the length of the warranty is 10.76 years.

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