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yaroslaw [1]
3 years ago
8

Lakisha has a total of $8000 to invest in two accounts for 1 year. One pays 5% simple interest, and the other pays 6% simple int

erest. How much should she invest in each account so the total interest earned is $438?
Mathematics
2 answers:
r-ruslan [8.4K]3 years ago
8 0
Let the amount invested in the first account be x and let the amount invested in the second be 8000 - x  
We have PRT/100 where P is her principal and r, her rate and t, the time 
So in the first account I = x * 5 * 1/100 
I = 5x/100 
On the other hand if she gets 6% interest in the second account, we have I = PRT/100 
So we have I = (8000 - x) * 6 * 1/ 100 
I = (48000 - 6x)/100 
But the combined interst is 438 so we have 
5x/100 + (48, 000 - 6x)/100 = 438 
Multiply through by 100 
5x + 48, 000 - 6x = 43800 
-x = 43800 - 48, 000 
-x = -4200 
Hence x = 4200. She will have to invest 8000 - 4200 in the second account which is 3800
expeople1 [14]3 years ago
5 0
Simple interest rate is given by:
I=(PRT)/100
let the amount invested in 5% S.I. be $x and amount invested in 6% S.I. be $(8000-x)
Interest obtained from 5% will be:
I=5/100×1×x=0.05x

Interest obtained from 6% will be:
I=6/100×1×(8000-x)=0.06(8000-x)
therefore total interest earned from the to schemes will be:
0.05x+0.06(8000-x)=438
solving for x we obtain:
0.05x+480-0.06x=438
0.05x-0.06x=438-480
-0.01x=-42
x=-42/(-0.01)
x=$4200
hence amount invested in 5% S.I. is $4200 and amount invested in 6% S.I. is (8000-4200)=$3800
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(a) The percent of her laps that are completed in less than 130 seconds is 55%.

(b) The fastest 3% of her laps are under 125.42 seconds.

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Step-by-step explanation:

The random variable <em>X</em> is defined as the number of seconds for a randomly selected lap.

The random variable <em>X </em>is normally distributed with mean, <em>μ</em> = 129.71 seconds and standard deviation, <em>σ</em> = 2.28 seconds.

Thus, X\sim N(129.71,\ 2.28^{2}).

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Compute the probability that a lap is completes in less than 130 seconds as follows:

P(X

                   =P(Z

The percentage is, 0.55 × 100 = 55%.

Thus, the percent of her laps that are completed in less than 130 seconds is 55%.

(b)

Let <em>x</em> represents the 3rd percentile.

That is, P (X < x) = 0.03.

⇒ P (Z < z) = 0.03

The value of <em>z</em> for the above probability is:

<em>z</em> = -1.88

Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sigma}\\-1.88=\frac{x-129.71}{2.28}\\x=129.71-(1.88\times 2.28)\\x=125.4236\\x\approx 125.42

Thus, the fastest 3% of her laps are under 125.42 seconds.

(c)

Let <em>x</em>₁ and <em>x</em>₂ be the values between which the middle 80% of the distribution lie.

That is,

P(x_{1}

The value of <em>z</em> for the above probability is:

<em>z</em> = 1.28

Compute the values of <em>x</em>₁ and <em>x</em>₂ as follows:

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