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ki77a [65]
3 years ago
8

$13,168is invested, part at 13%and the rest at

Mathematics
1 answer:
Kryger [21]3 years ago
7 0

Answer: the amount invested at 13% is $9748

the amount invested at 5% is $3420

Step-by-step explanation:

Let x represent the amount invested at 13%.

Let y represent the amount invested at 5%.

$13,168 is invested, part at 13 % and the rest at 5%. This means that

x + y = 13168

The formula for determining simple interest is expressed as

I = PRT/100

Where

I represents interest paid on the loan.

P represents the principal or amount taken as loan

R represents interest rate

T represents the duration of the loan in years.

Considering the amount invested at 13%,

P = x

R = 13%

T =1 year

I = (x × 13 × 1)/100 = 0.13x

Considering the amount invested at 5%,

P = y

R = 5%

T = 1 year

I = (x × 5 × 1)/100 = 0.05y

If the interest earned from the amount invested at

13% exceeds the interest earned from the amount invested at

5% by $1438.24, it means that

0.13x - 0.05y = 1438.24 - - -- - - - - -1

Substituting x = 13168 - y into equation 1, it becomes

0.13(13168 - y) - 0.05y = 1438.24

1711.84 - 0.13y - 0.05y = 1438.24

- 0.13y - 0.05y = 1438.24 - 1711.84

- 0.08y = - 273.6

y = - 273.6/- 0.08

y = 3420

x = 13168 - y = 13168 - 3420

x = 9748

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Answer:

Answer is B

Step-by-step explanation:

6 0
3 years ago
Lucía makes braided key chains out of cotton cord. For every 8 feet of cotton cord she has, she can make 3 braided key chains. P
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Lucía needs 13.33 feet of cotton.

Since Lucía makes braided key chains out of cotton cord, and for every 8 feet of cotton cord she has, she can make 3 braided key chains, to represent the relationship between the feet of cotton cord and the number of key chains and find the number of feet of cotton cord Lucia needs to make 5 key chains, the following calculations must be performed, through a cross multiplication:

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Therefore, Lucía needs 13.33 feet of cotton.

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7 0
2 years ago
The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard dev
Vladimir [108]

Answer:

Probability that the average length of a sheet is between 30.25 and 30.35 inches long is 0.0214 .

Step-by-step explanation:

We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches.

Also, a sample of four metal sheets is randomly selected from a batch.

Let X bar = Average length of a sheet

The z score probability distribution for average length is given by;

                Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 30.05 inches

           \sigma   = standard deviation = 0.2 inches

             n = sample of sheets = 4

So, Probability that average length of a sheet is between 30.25 and 30.35 inches long is given by = P(30.25 inches < X bar < 30.35 inches)

P(30.25 inches < X bar < 30.35 inches)  = P(X bar < 30.35) - P(X bar <= 30.25)

P(X bar < 30.35) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{30.35-30.05}{\frac{0.2}{\sqrt{4} } } ) = P(Z < 3) = 0.99865

 P(X bar <= 30.25) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{30.25-30.05}{\frac{0.2}{\sqrt{4} } } ) = P(Z <= 2) = 0.97725

Therefore, P(30.25 inches < X bar < 30.35 inches)  = 0.99865 - 0.97725

                                                                                       = 0.0214

                                       

7 0
3 years ago
Which shows 24+54 written using GCF and the distributive property
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