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3 years ago

A

Mathematics

2 answers:

fredd [130]3 years ago

7 0

The answer to your question 2456

salantis [7]3 years ago

4 0

Answer:

54280

Step-by-step explanation:

We are given a lot of information here, so the first thing we need to do is figure out what the problem is. We are asked to find the amount left in the IRA if the retiree takes money out now. To find this, all we need to know is his age, the amount in his IRA account, the percent he plans to take out, the age where no penalty is taken for withdrawals, and the penalty percentage for taking out money before age is not reached. All the other information is irrelevant. We are told that he is 68 and the age for no penalty is 72. Now we need the percent he wants to take out, which is 2% and the penalty is 6%. We are also told he has $59,000 in his IRA account. We first add the percent he plans to take out with the penalty percent, then multiply by the amount in his IRA

2+6=8%

0.08⋅59000=4720

Now, we subtract that amount from the original IRA amount

59000−4720=$54,280

So, $54,280 will be left in his IRA.

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Is this pattern a net for the three-dimensional figure?<br> Yes<br> No

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3 years ago

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Solve for v.<br> 3v = 42<br> Simplify your answer as much as possible.

pogonyaev

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2 years ago

A circular rug has a diameter of 6 feet. What is the area of the rug?

podryga [215]

Answer:

The answer to your question is: letter B

Step-by-step explanation:

Data

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3 years ago

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The length of the base of an isosceles triangle is x. The length of a leg is 2x-6. The perimeter of the triangle is 38. Find X

Norma-Jean [14]

Answer:

x = 10

Step-by-step explanation:

Perimeter of a triangle = the sum of the three sides.

In an isosceles the legs are equal (2x - 6) and the base is x

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3 years ago

Although cities encourage carpooling to reduce traffic congestion, most vehicles carry only one person. For example, 64% of vehi

ira [324]

Answer:

a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

b) 0.996 is the probability that more than half of the vehicles carry just one person.

Step-by-step explanation:

We are given the following information:

A) Binomial distribution

We treat vehicle on road with one passenger as a success.

P(success) = 64% = 0.64

Then the number of vehicles follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

We have to evaluate:

$P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291$

0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

B) By normal approximation

Sample size, n = 92

p = 0.64

$\mu = np = 92(0.64) = 58.88$

$\sigma = \sqrt{np(1-p)} = \sqrt{92(0.64)(1-0.64)} = 4.60$

We have to evaluate the probability that more than 47 cars carry just one person.

$P(x \geq 47)$

After continuity correction, we will evaluate

$P( x \geq 46.5) = P( z > \displaystyle\frac{46.5 - 58.88}{4.60}) = P(z > -2.6913)$

$= 1 - P(z \leq -2.6913)$

Calculation the value from standard normal z table, we have,

$P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%$

0.996 is the probability that more than half out of 92 vehicles carry just one person.

5 0

3 years ago