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Len [333]
3 years ago
6

the description of a TV is based on the diagonal length. if the height of a TV is 10 inches and the width is 7 inches, what is t

he approximate diagonal length in inches?
Mathematics
1 answer:
borishaifa [10]3 years ago
8 0
To calculate it use pythagoras: a^2+b^2=c^2
7^2+10^2=49+100=149

calculate the square root: \sqrt{149}=12.2
so the diagonal is about 12.2 inches long

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What is the distance and midpoint of (-2,-2) and (3,5)​
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2 years ago
Chin-li is building a brick wall along the front of his property. The wall will have 15 rows of bricks, with 32 bricks in each r
Finger [1]

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<h3>The solution is 480 bricks.</h3>

Step-by-step explanation:

We are given that number of rows of bricks in a wall = 15 rows.

Number of bricks in a row = 32 bricks.

In order to find the total number of bricks, we need to multiply number of rows with number of bricks in a row.

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The number of major earthquakes in a year is approximately normally distributed with a mean of 20.8 and a standard deviation of
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Answer:

a) 51.60% probability that in a given year there will be less than 21 earthquakes.

b) 49.35% probability that in a given year there will be between 18 and 23 earthquakes.

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:

\mu = 20.8, \sigma = 4.5

a) Find the probability that in a given year there will be less than 21 earthquakes.

This is the pvalue of Z when X = 21. So

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

Z = \frac{21 - 20.8}{4.5}

Z = 0.04

Z = 0.04 has a pvalue of 0.5160.

So there is a 51.60% probability that in a given year there will be less than 21 earthquakes.

b) Find the probability that in a given year there will be between 18 and 23 earthquakes.

This is the pvalue of Z when X = 23 subtracted by the pvalue of Z when X = 18. So:

X = 23

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

Z = \frac{23 - 20.8}{4.5}

Z = 0.71

Z = 0.71 has a pvalue of 0.7611

X = 18

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

Z = \frac{18 - 20.8}{4.5}

Z = -0.62

Z = -0.62 has a pvalue of 0.2676

So there is a 0.7611 - 0.2676 = 0.4935 = 49.35% probability that in a given year there will be between 18 and 23 earthquakes.

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