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Alexxx [7]
3 years ago
11

What is the answer of how much he got in interest?

Mathematics
2 answers:
beks73 [17]3 years ago
8 0

Answer:

He earned $1,100 in interest. $6,600 is the new amount in his bank.

Step-by-step explanation:

Liula [17]3 years ago
5 0

Answer:

After 4 he will have earned $1100 in Interest

His new account balance will be $6600

Step-by-step explanation:

Principal P = $5500, Time T = 4 years, and at Rate R of 5%

Simple Interest SI = PRT/100

SI = (5500 × 4 × 5)/100

SI = 55 × 4 × 5

SI =  $1100

Account balance after 4 years will be Principal + Simple Interest

$5500 + $1100 = $6600

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PLEASE HELP!!! Which expression is equivelent to 32 + 16 1. 8(4+2) 2. 8(24+8) 3. 4(2+7) 4. 7(24+8)
victus00 [196]

Answer is

1. 8(4+2)

8 × 6 = 48

6 0
3 years ago
Create a set of data with 7 values that has a mean of 30, a median of 26, a range of 50, and an interquartile range of 36.
kotykmax [81]

5, 10, 23, 26, 45, 46, 55

7 0
3 years ago
WITH ATTACHMENT 1. Find the length of the midsegment. The diagram is not to scale.(1 point)
LUCKY_DIMON [66]
From the given attachment, the triangles are similar.
45/(6x + 4) = 90/(4x + 56)
45(4x + 56) = 90(6x + 4)
180(x + 14) = 180(3x + 2)
3x - x = 14 - 2
2x = 12
x = 12/2 = 6

Length of midsegment = 6x + 4 = 6(6) + 4 = 36 + 4 = 40
8 0
3 years ago
Read 2 more answers
Betsy made ribbons for school spirit day. Her roll was 30 ft. Long. For each individual ribbon she needed 0.625 ft. How many rib
Juli2301 [7.4K]

Divide the total amount of ribbon by the length of each individual ribbon:

30 ft / 0.625 ft = 48

She can make 48 ribbons.

5 0
4 years ago
In a large company, the proportion of employees who were promoted during the last year was 0.10. If 100
Natasha2012 [34]

Answer:

0.0475

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed distributions 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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In a large company, the proportion of employees who were promoted during the last year was 0.10.

This means that p = 0.1

100 employees

This means that n = 100

Mean and standard deviation:

\mu = E(X) = np = 100*0.1 = 10

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.1*0.9} = 3

What is the probability that at least 15 of them were promoted during the last year?

This is P(X \geq 15), which is 1 subtracted by the pvalue of Z when X = 15. So

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

Z = \frac{15 - 10}{3}

Z = 1.67

Z = 1.67 has a pvalue of 0.9525

1 - 0.9525 = 0.0475.

0.0475 is the answer.

6 0
3 years ago
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