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Peter has money in two savings accounts. One rate is 6% and the other is 14%. If he has $900 more in the 14% account and the tot

-BARSIC- [3]

Answer:

$1375 at 14%
$475 at 6%

Step-by-step explanation:

Let x represent the amount Peter invested at 14%. Then (x-900) is the amount he invested at 6%. His total interest earned is ...

0.14x + 0.06(x -900) = 221

0.20x = 275 . . . . . . . . . . . . . . . . add 54, simplify

x = 1375 . . . . . . . . . . . . . . . . . . . .divide by 0.2; amount invested at 14%

(x-900) = 475 . . . . . . . . . . . . . . . amount invested at 6%

Peter invested $475 in the 6% account and $1375 in the 14% account.

7 0

3 years ago

WILL GIVE BRAINLIEST PLEASE HELP

Debora [2.8K]

Answer:

the 3 one is the correct answer

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

8.52 The heights of 2-year-old children are normally distributed with a mean of 32 inches and a standard deviation of 1.5 inches

sweet [91]

Answer:

Heights of 29.5 and below could be a problem.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

The heights of 2-year-old children are normally distributed with a mean of 32 inches and a standard deviation of 1.5 inches.

This means that $\mu = 32, \sigma = 1.5$