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Lisa [10]
3 years ago
7

You deposit $300 into a savings account that pays 4% interest compounded annually. How much interest will you earn after 5 years

?
Question 4 options:

$65


$60


$360


$365
Mathematics
1 answer:
sammy [17]3 years ago
4 0

Answer:

365

Step-by-step explanation:

300X1.04^5=364.995871

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Reil [10]
I hope this helps you




4r=p-3t



r=p-3t/4
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3 years ago
Please help me solve this practice SAT question!
lord [1]

Answer:

t=5

Step-by-step explanation:

∵ (t,17) lies on (x,y) plane y=x²-3x+7,t>0

∴ 17=t²-3t+7

t²-3t+7-17=0

t²-3t-10=0

t²-5t+2t-10=0

t(t-5)+2(t-5)=0

(t-5)(t+2)=0

t=5,-2

t>0

so t=5

6 0
3 years ago
In circle C, what is the value of X?
solong [7]
The answer should be 112 degrees
7 0
4 years ago
Read 2 more answers
Movies are on sale for 3/4 the original price. If the original price is $10, what is the sale price?
gtnhenbr [62]
3/4 of 10 would = $7.5 which would be your answer.
3 0
4 years ago
In a string of 12 Christmas tree light bulbs, 3 are defective. The bulbs are selected at random and tested, one at a time, until
Juliette [100K]

Using the hypergeometric distribution, it is found that there is a 0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

In this problem, the bulbs are chosen without replacement, hence the <em>hypergeometric distribution</em> is used to solve this question.

<h3>What is the hypergeometric distribution formula?</h3>

The formula is:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • N is the size of the population.
  • n is the size of the sample.
  • k is the total number of desired outcomes.

In this problem:

  • There are 12 bulbs, hence N = 12.
  • 3 are defective, hence k = 3.

The third defective bulb is the fifth bulb if:

  • Two of the first 4 bulbs are defective, which is P(X = 2) when n = 4.
  • The fifth is defective, with probability of 1/8, as of the eight remaining bulbs, one will be defective.

Hence:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,12,4,3) = \frac{C_{3,2}C_{9,1}}{C_{12,4}} = 0.2182

0.2182 x 1/8 = 0.0273.

0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

More can be learned about the hypergeometric distribution at brainly.com/question/24826394

8 0
2 years ago
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