To what amount will $1500 grow if compounded semiannually at 6.75% interest for 10 years?

The local hardware store has blue buckets that hold 2 gallons of water and white buckets that hold 5 gallons of water. You bough

Katarina [22]

Answer:

4 White, 3 Blue

Step-by-step explanation:

5w+2b=26

Closest five multiple that leaves an even remainder is 4.

5×4=20

26-20=6÷2=3

5 0

4 years ago

Read 2 more answers

A manufacturer reported a sample mean = 22.0 g and a sample standard deviation = 2.5 g based on a sample of 20 of their products

tester [92]

Answer:

$n=(\frac{1.640(2.5)}{2})^2 =4.2 \approx 5$

So the answer for this case would be n=5 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=22$ represent the sample mean

$\mu$ population mean (variable of interest)

s=2.5 represent the sample standard deviation

n represent the sample size

ME = 2 represent the margin of error accepted

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (2)

And on this case we have that ME =2 and we are interested in order to find the value of n, if we solve n from equation (2) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (3)

We can use as estimator for the population deviation the sample deviation $\hat \sigma = s$

The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.05;0;1)", and we got $z_{\alpha/2}=1.640$ , replacing into formula (3) we got:

$n=(\frac{1.640(2.5)}{2})^2 =4.2 \approx 5$

So the answer for this case would be n=5 rounded up to the nearest integer

6 0

3 years ago

The random variable X, representing the number of accidents in a certain intersection in a week, has the following probability d

ollegr [7]

Answer: 0.70

Step-by-step explanation:

Given : The random variable X, representing the number of accidents in a certain intersection in a week, has the following probability distribution:

x 0 1 2 3 4 5

P(X = x) 0.20 0.30 0.20 0.15 0.10 0.05

Using the above probability distribution , the the probability that in a given week there will be at most 3 accidents is given by :_

$P(\leq3)=P(0)+P(1)+P(2)+P(3)\\\\=0.20+0.30+0.20=0.70$