V=(1/3)hpir^2
v=130
h=13
r=x
130=(1/3)(13)pix^2
times both sides by 3
390=13pix^2
divide both sides by 13
30=pix^2
divide both sides by pi
30/pi=x^2
sqrt both sides
x≈3.09
radius is abou 3 units
Answer:
is in proportion.
is not in proportion.
is in proportion.
is in proportion.
Step-by-step explanation:
The first example is
and it is in proportion.
This is because, you will get the same simplest fraction (
) from the fraction
by dividing its numerator and denominator by 5.
The second example is
and it is not in proportion.
This is because, you can not get the simplest fraction (
) from the fraction
after simplification.
The third example is
and it is in proportion.
This is because, you will get the same simplest fraction (
) from the fraction
by dividing its numerator and denominator by 2 and from the fraction
by dividing its numerator and denominator by 3.
The fourth example is
and it is in proportion.
This is because, you will get the same simplest fraction (
) from the fraction
by dividing its numerator and denominator by 5. (Answer)
Answer:
112.5 which is 113
Step-by-step explanation:
150 divided by 60= 2.5
2.5 times 45=112.5
Using the normal distribution, it is found that there is a 0.2776 = 27.76% probability that the life span of the monitor will be more than 20,179 hours.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:

- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:

The probability that the life span of the monitor will be more than 20,179 hours is <u>one subtracted by the p-value of Z when X = 20179</u>, hence:


Z = 0.59.
Z = 0.59 has a p-value of 0.7224.
1 - 0.7224 = 0.2776.
0.2776 = 27.76% probability that the life span of the monitor will be more than 20,179 hours.
More can be learned about the normal distribution at brainly.com/question/24663213
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