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ANTONII [103]
3 years ago
5

Need help! plz!!!!!!!

Mathematics
1 answer:
Dovator [93]3 years ago
6 0
If the figure is drawn in a 2/5 scale, this means that all the length will be multiplied by 2/5. The area is a two-dimensional thing, so it would be multiplied by that factor twice, or 4/25. You can try checking all the lengths. The correct one should match all of the things I said above.

The answer is the top left one.
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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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3 years ago
Using the ratio 2 out of every 5 students
Hatshy [7]

Answer:

20

Step-by-step explanation:

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3 years ago
!!!WILL MARK BRAINLIEST!!
Leya [2.2K]

Answer:

the answer is D

Step-by-step explanation:

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3 years ago
Describe how the graph of the parent function y=%x
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Answer:

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3 years ago
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Find the value of each variable in the circle to the right. The dot represents the center of the circle.
MrMuchimi

9514 1404 393

Answer:

  a = 57.5

  b = 90

  c = 65

Step-by-step explanation:

The measure of an inscribed angle is half that of the arc it intercepts. The angle at 'a' intercepts an arc of 115°, so its measure is (115/2)° = 57.5°

__

The inscribed angle at 'b' is a semicircle, 180°, so the measure of 'b' is (180/2)° = 90°.

__

The arc 'c' is the remainder of a semicircle after 115° have been taken away. Its measure is 180° -115° = 65°.

  (a, b, c) = (57.5, 90, 65)

4 0
3 years ago
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