Nickels is 0,05 of dolar - n
dimes is 0,1 o dolar -d
quarters is 0,25 of dolar -q
dimes =2nickels
d=2n
quarters=2dimes
q=2d=4n
12,5=n(0,05)+2n(0,1)+4n(0,25)
12,5=0,05n+0,2n+n
12,5=1,25n
n=10
q=4n=40
There are 40 quarters

5 0

3 years ago

SIMPLYFY THE EXPRESSION WRITE THE ANSWER AS A POWER (5^4)^3

Anastaziya [24]

Answer:

5^12

Step-by-step explanation:

You multiply both roots

5 0

2 years ago

An environmentalist wants to find out the fraction of oil tankers that have spills each month. Step 2 of 2 : Suppose a sample of

rewona [7]

Answer:

The 80% confidence interval for the population proportion of oil tankers that have spills each month is (0.199, 0.257).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Suppose a sample of 333 tankers is drawn. Of these ships, 257 did not have spills.

333 - 257 = 76 have spills.

This means that $n = 333, \pi = \frac{76}{333} = 0.228$

80% confidence level

So $\alpha = 0.2$ , z is the value of Z that has a pvalue of $1 - \frac{0.2}{2} = 0.9$ , so $Z = 1.28$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.228 - 1.28\sqrt{\frac{0.228*0.772}{333}} = 0.199$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.228 + 1.28\sqrt{\frac{0.228*0.772}{333}} = 0.257$

The 80% confidence interval for the population proportion of oil tankers that have spills each month is (0.199, 0.257).

6 0

3 years ago