Find the Geometric mean x of 5 and 20

The sum of 3x^2+4x-2 and x^2-5x+3 is

Diano4ka-milaya [45]

The sum would be

4x^2-x+1

add like terms.

3x^2+x^2

4x^2

4x+-5x

-x

3-2

1

3 0

3 years ago

Read 2 more answers

Bob Nale is the owner of Nale’s Quick Fill. Bob would like to estimate the mean number of gallons of gasoline sold to his custom

ira [324]

Answer:

a) The best estimator for the population mean is given by the sample mean $\hat \mu = \bar X = 5.70$

b) The 99% confidence interval would be given by (4.969;6.431)

c) We are 99% confident that the true mean for th number of gallons of gasoline sold to his customers is between 4.969 and 6.431.

Step-by-step explanation:

1) 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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=1.9$ represent the population standard deviation

n=45 represent the sample size

2) Part a

The best estimator for the population mean is given by the sample mean $\hat \mu = \bar X = 5.70$

3) Part b. Develop a 99 percent confidence interval for the population mean.

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 $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

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

$5.7-2.58\frac{1.9}{\sqrt{45}}=4.969$

$5.7+2.58\frac{1.9}{\sqrt{45}}=6.431$

So on this case the 99% confidence interval would be given by (4.969;6.431)

4) Part c. Interpret the meaning of part (b).

On this case we can say this: We are 99% confident that the true mean for th number of gallons of gasoline sold to his customers is between 4.969 and 6.431.

4 0

3 years ago

It takes a hose 2 minutes to fill a rectangular aquarium 8 inches long, 9 inches wide, and 13 inches tall. How long will it take

nirvana33 [79]

To make it easier, you calculate the volume of the first aquarium.

1st aquarium:
V = L x W x H
V = 8 x 9 x 13
V = 72 x 13
V = 936 in.
Rate: 936 in./2 min.

Now that you've got the volume and rate of the first aquarium, you can find how many inches of the aquarium is filled within a minute, which is also known as the unit rate. To do that, you have to divide both the numerator and denominator by their least common multiple, which is 2. 936 divided by 2 is 468 and 2 divided by 2 is 1.
So the unit rate is 468 in./1 min. Now that you've got the unit rate, you can find out how long it'll take to fill the second aquarium up by finding its volume first.

2nd aquarium:
V = L x W x H
V = 21 x 29 x 30
V = 609 x 30
V 18,270 inches

Calculations:
Now, you divide 18,270 by 468 to find how many minutes it will take to fill up the second aquarium. 18,270 divided by 468 is about 39 (the answer wasn't exact, so I said "about").

2nd aquarium's rate:
18,270 in./39 min.

As a result, it'll take about 39 minutes to fill up an aquarium measuring 21 inches by 29 inches by 30 inches using the same hose. I really hope I helped and that you understood my explanation! :) If I didn't, I'm sorry. I tried. :(

7 0

3 years ago

What is the partial products that results from multiplying 15 times 32

sergejj [24]

See attached PDF. (The censor thinks there are some unseemly words in there.)

Download pdf

6 0

3 years ago

) You are finally well!  But now your child is sick!  You know there have been mistakes in orders so you are on the internet t

rjkz [21]

Answer:

The doctor's dosage was not appropriate.