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Nina [5.8K]
3 years ago
14

Ms. Wilson invested ​$34000 in two​ accounts, one yielding 8​% interest and the other yielding 9​%. If she received a total of​

$2910 in interest at the end of the​ year, how much did she invest in each​ account?
Mathematics
1 answer:
Digiron [165]3 years ago
3 0

Answer:

Amount invested at 8 % rate = x = $ 15000

Amount invested at 9 % rate = 34000 - x = 34000 - 15000 = $ 19000

Step-by-step explanation:

Total Amount = $ 34000

Let amount invested at 8 % rate = x

Amount invested at 9 % rate = $ 34000 - x

Total interest = $ 2910

2910 = \frac{x (8) (1)}{100} + \frac{(34000-x) (9) (1)}{100}

291000 = 8 x + 306000 - 9 x

x = 306000 - 291000

x = 15000

So  amount invested at 8 % rate = x = $ 15000

Amount invested at 9 % rate = 34000 - x = 34000 - 15000 = $ 19000

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You always need some time to get up after the alarm has rung. You get up from 10 to 20 minutes later, with any time in that inte
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Answer:

a) P(x<5)=0.

b) E(X)=15.

c) P(8<x<13)=0.3.

d) P=0.216.

e) P=1.

Step-by-step explanation:

We have the function:

f(x)=\left \{ {{\frac{1}{10},\, \, \, 10\leq x\leq 20 } \atop {0, \, \, \, \, \, \,  otherwise }} \right.

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c) We calculate  the probability that you will need between 8 and 13 minutes:

P(8\leq x\leq 13)=P(10\leqx\leq 13)\\\\P(8\leq x\leq 13)=\int_{10}^{13} f(x)\, dx\\\\P(8\leq x\leq 13)=\int_{10}^{13} \frac{1}{10} \, dx\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot [x]_{10}^{13}\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot (13-10)\\\\P(8\leq x\leq 13)=\frac{3}{10}\\\\P(8\leq x\leq 13)=0.3

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d)  We calculate the probability that you will be late to each of the 9:30am classes next week:

P(x>14)=\int_{14}^{20} f(x)\, dx\\\\P(x>14)=\int_{14}^{20} \frac{1}{10} \, dx\\\\P(x>14)=\frac{1}{10} [x]_{14}^{20}\\\\P(x>14)=\frac{6}{10}\\\\P(x>14)=0.6

You have 9:30am classes three times a week.  So, we get:

P=0.6^3=0.216

Therefore, the probability is P=0.216.

e)  We calculate the probability that you are late to at least one 9am class next week:

P(x>9.5)=\int_{10}^{20} f(x)\, dx\\\\P(x>9.5)=\int_{10}^{20} \frac{1}{10} \, dx\\\\P(x>9.5)=\frac{1}{10} [x]_{10}^{20}\\\\P(x>9.5)=1

Therefore, the probability is P=1.

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