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Papessa [141]
2 years ago
14

Two-thirds of a number is nineteen.

Mathematics
2 answers:
Brilliant_brown [7]2 years ago
7 0

Answer:

28.5

Step-by-step explanation:

X*2/3=19, X=19/(2/3)=57/2=28.5

mart [117]2 years ago
6 0

Answer:

The number is 28.5

Step-by-step explanation:

Let the number be x

Two - third of the number is 19 means

                                                            \frac{2}{3} \ of \ x = 19

<em><u>Solve for x</u></em>

<em><u></u></em>\frac{2}{3} \times x = 19\\\\2 \times x = 19 \times 3\\\\x = \frac{19 \times 3}{2} = 28.5<em><u></u></em>

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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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