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3 years ago

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What is m∠S? Enter your answer in the box. ° An isosceles triangle with vertices labeled R, S, and T. Side R T is the base. Side

s R S and side S T are equal. Angle R is labeled as 3 x degrees, angle S is labeled as 2 x degrees, and angle T is labeled as 3 x degrees.

1 answer:

yulyashka [42]3 years ago

5 0

Answer:

m<S = 45°

Step-by-step explanation:

The sum of the measures of the angles of a triangle equals 180 deg.

m<R + m<S + m<T = 180

3x + 2x + 3x = 180

8x = 180

x = 22.5

m<S = 2x = 2(22.5) = 45

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2 years ago

Question Help Suppose that the lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a

koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

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

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So $\mu = 5656, \sigma = 333.3$

(a) What proportion of light bulbs will last more than 6262 hours?

The pvalue of the z-score of $X = 6262$ is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

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

$Z = \frac{6262 - 5656}{333.3}$

$Z = 1.82$

$Z = 1.81$ has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

$P = 100% - 96.562% = 3.438%$ of the light bulbs will last more than 6262 hours.

(b) What proportion of light bulbs will last 5252 hours or less?

This is the pvalue of the zscore of $X = 5252$

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

$Z = \frac{5252- 5656}{333.3}$

$Z = -1.21$

$Z = -1.21$ has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 hours?

This is the pvalue of the zscore of $X = 6262$ subtracted by the pvalue of the zscore $X = 5858$

For $X = 6262$ , we have that $Z = 1.81$ with a pvalue of .96562.

For X = 5858

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

$Z = \frac{5858- 5656}{333.3}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

$P = .96562 - .72907 = .23655$

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

(d) What is the probability that a randomly selected light bulb lasts less than 4646 hours?

This is the pvalue of the zscore of $X = 4646$

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

$Z = \frac{4646- 5656}{333.3}$

$Z = -3.03$

$Z = -3.03$ has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

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2 years ago

Find in degrees.<br> 11<br> B<br> 6<br> [?]°<br> Round to the nearest hundredth.<br> Enter

avanturin [10]

Answer:

$\beta = 28.610$

Step-by-step explanation:

in this case we have a right angled triangle, whenever you see this triangle think of the TRIG RATIOS * sin, cos & tan* but if it was another triangle besides right angled triangle you use SINE RULE OR COSINE RULE

SOLUTION

$\tan( \beta ) = \frac{opposite}{adjecent} \\ \tan( \beta ) = \frac{6}{11} \\ \beta = tan(inverse)( \frac{6}{11} ) \\ \beta = 28.61045$

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