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How many quarters can you get for a five dollar bill

sweet [91]

You can get 20 quarters for a five dollar bill. Just divide 5.00 by 0.25 and you get 20. To check, you can multiply 20 by 0.25 and you will get 5.00 or 5.

4 0

3 years ago

find an nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing utili

Amiraneli [1.4K]

Since 4i is a root, that automatically means -4i is also a root (complex roots always travel in pairs) So the factored form would look like (x+4i)(x−4i). When multiplied out this gives the polynomial x2+16.

5 0

3 years ago

Greg goes fishing every day for a week. On the first day he catches seven fish and each

Oliga [24]

so first day and so on

7, 10, 13,....

as you can see it's an arithmetic progression

so sum for nth term= n/2 { 2a + (n-1) d}

it's the sum of the 7th term

so

7/2 { 7 ×2 + ( 7-1) 3}

7/2 × 32

7× 16

112 fishes

6 0

3 years ago

Read 2 more answers

GRE Test: The Graduate Record Examination (GRE) is a test required for admission to many US graduate schools. Student’s scores o

Zepler [3.9K]

Answer:

a) $\bar X=144.3$

b) The 95% confidence interval would be given by (138.846;149.754)

c) n=34

d) n=298

e) n=1190

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".

We can calculate the sample mean with this formula:

$\bar X = \frac{\sum_{i=1}^n x_i}{n}$

$\bar X=144.3$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=8.8$ represent the population standard deviation

n=10 represent the sample size

a) Find a point estimate of the population mean

We can calculate the sample mean with this formula:

$\bar X = \frac{\sum_{i=1}^n x_i}{n}$

$\bar X=144.3$

b) Construct a 95% confidence interval for the true mean score for this population.

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

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

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $t_{\alpha/2}=1.96$

Now we have everything in order to replace into formula (1):

$144.3-1.96\frac{8.8}{\sqrt{10}}=138.846$

$144.3+1.96\frac{8.8}{\sqrt{10}}=149.754$

So on this case the 95% confidence interval would be given by (138.846;149.754)

Part c

The margin of error is given by this formula:

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

And on this case we have that ME =+20 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

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

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , replacing into formula (5) we got: