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PolarNik [594]
3 years ago
13

The amount of jen's monthly phone bill is normally distributed with a mean of $65 and a standard deviation of 11. what percentag

e of her phone bills are between 32 and 98
Mathematics
1 answer:
stepan [7]3 years ago
4 0
It would be 99%. If you draw the normal distribution curve and plot the mean and standard deviation, it might help you visually.

:)



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Based on the work shown on the right, check all of the possible solutions of the equation.
Mazyrski [523]

Incomplete question. However, here is a similar question attached.

Solve 3x^2 + 17x - 6 = 0.

Based on the work shown to the left, which of these values are possible solutions of the equation? Check all of the boxes that apply

A X=-6

B X=6

C X=-1/3

D X=1/3

E X=0

Answer:

A and D

Step-by-step explanation:

Note that such question requires using completing the square method of solving equations. By using the values X= -6 and X= 1/3 we arrive at a solution.

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3 years ago
A study suggested that childrenbetween the ages of 6 and 11 in the US have anaverage weightof 74 lbs. with a standard deviation
Doss [256]

Answer:

The proportion of children in this age range between 70 lbs and 85 lbs is of 0.9306.

Step-by-step explanation:

When the distribution is normal, we use 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.

A study suggested that children between the ages of 6 and 11 in the US have an average weightof 74 lbs, with a standard deviation of 2.7 lbs.

This means that \mu = 74, \sigma = 2.7

What proportion of childrenin this age range between 70 lbs and 85 lbs.

This is the pvalue of Z when X = 85 subtracted by the pvalue of Z when X = 70. So

X = 85

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

Z = \frac{85 - 74}{2.7}

Z = 4.07

Z = 4.07 has a pvalue of 1

X = 70

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

Z = \frac{70 - 74}{2.7}

Z = -1.48

Z = -1.48 has a pvalue of 0.0694

1 - 0.0694 = 0.9306

The proportion of children in this age range between 70 lbs and 85 lbs is of 0.9306.

7 0
2 years ago
PLEASE HELP MATH
Novosadov [1.4K]
I believe C is your answer. 
I hope this helps!
7 0
2 years ago
Read 2 more answers
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