I have no clue what to do here:

How much must be deposited today into the following account in order to have $ 65,000 in 8 years for a down payment on a house?

svp [43]

Answer:

$\$47,494.86$

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=8\ years\\ A=\$65,000\\ r=4\%=4/100=0.04\\n=1$

substitute in the formula above

$65,000=P(1+\frac{0.04}{1})^{1*8}$

$65,000=P(1.04)^{8}$

Solve for P

$P=65,000/(1.04)^{8}$

$P=\$47,494.86$

3 0

3 years ago

$1500.00 in a simple interest savings account. Two years later, the total account balance is $1665.00.

antoniya [11.8K]

Answer:

interest rate=45.04

I after 8 years=5994

step on step explainaton:

I=PRT

1500=1665×r/100×2

r=45.04

I after 8 years= I=PRT

=1665×45/100×8=5994

6 0

4 years ago

Assume that females have pulse rates that are normally distributed with a mean of mu equals 72.0 beats per minute and a standard

larisa86 [58]

Answer:

a) 36.88% probability that her pulse rate is between 66 beats per minute and 78 beats per minute

b) 66.30% probability that they have pulse rates with a mean between 66 beats per minute and 78 beats per minute

c)

C.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 72, \sigma = 6.5$

a. If 1 adult female is randomly selected, find the probability that her pulse rate is between 66 beats per minute and 78 beats per minute.

The probability is?

This is the pvalue of Z when X = 78 subtracted by the pvalue of Z when X = 66. So

X = 78

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{78 - 72}{12.5}$

$Z = 0.48$

$Z = 0.48$ has a pvalue of 0.6844

X = 66

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{66 - 72}{12.5}$

$Z = -0.48$

$Z = -0.48$ has a pvalue of 0.3156

0.6844 - 0.3156 = 0.3688

36.88% probability that her pulse rate is between 66 beats per minute and 78 beats per minute

b. If 4 adult females are randomly selected, find the probability that they have pulse rates with a mean between 66 beats per minute and 78 beats per minute

The probability is?

Now we have $n = 4, s = \frac{12.5}{\sqrt{4}} = 6.25$

So

X = 78

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{78 - 72}{6.25}$

$Z = 0.96$

$Z = 0.96$ has a pvalue of 0.8315

X = 66

$Z = \frac{X - \mu}{s}$

$Z = \frac{66 - 72}{6.25}$

$Z = -0.96$

$Z = -0.96$ has a pvalue of 0.1685

0.8315 - 0.1685 = 0.6630

66.30% probability that they have pulse rates with a mean between 66 beats per minute and 78 beats per minute

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

The condition for the sample size exceeding 30 is when the population is skewed. If it is normally distributed, the size is not a condition.

So the correct answer is:

C.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

3 0

3 years ago

Read 2 more answers

Write a multiplication story for 4x9 then find the product

skelet666 [1.2K]

Answer:

You are setting up chairs for the band concert tonight, the band conductor tells you to set up four rows with nine chairs each. How many chairs will there be all together.

to solve:

4 x 9 = 36

or 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 = 36

7 0

4 years ago

Read 2 more answers

Let f(x)=3x+5 and g(x)=x^2.

Nady [450]

Answer: x² + 3x + 5

<u>Step-by-step explanation:</u>

f(x) = 3x + 5 g(x) = x²

(f + g)(x) = f(x) + g(x)

= 3x + 5 + x²

= x² + 3x + 5

3 0

3 years ago