Find all the numbers such that one-third of the number is at most five.

Solve (16×^3-3×^2+43×-5)÷(×^2-2)

Oksana_A [137]

Answer:

3872/(x^2 - 2)

Step-by-step explanation:

Simplify the following:

(16^3 - 3^2 - 5×43)/(x^2 - 2)

3^2 = 9:

(16^3 - 9 - 5×43)/(x^2 - 2)

16^3 = 16×16^2:

(16×16^2 - 9 - 5×43)/(x^2 - 2)

| 1 | 6

× | 1 | 6

| 9 | 6

1 | 6 | 0

2 | 5 | 6:

(16×256 - 9 - 5×43)/(x^2 - 2)

16×256 = 4096:

(4096 - 9 - 5×43)/(x^2 - 2)

43 (-5) = -215:

(4096 - 9 + -215)/(x^2 - 2)

4096 - 9 - 215 = 3872:

Answer: 3872/(x^2 - 2)

7 0

3 years ago

Cindy has 26 nickels. She is getting rolls of nickels from the bank. She has enough money to get up to 10 rolls of nickels and e

Alchen [17]

We know that the domain is actually the number of rolls that she gets because it is the input variable in the equation f (r) = 40 r + 26. The number of rolls that she can get can never be a negative number nor a zero nor a decimal number. However she can only get up to 10 rolls. Therefore the answer is:

all integers from 1 to 10, inclusive

8 0

3 years ago

How do you do this? can someone explain it to me please?

My name is Ann [436]

Answer:

you have to detect the angles on itmeasure the angles on the empty side wheree there are no numbers based on what the other side is

Step-by-step explanation:

7 0

3 years ago

F there are 10 yellow Skittles, how many are green?

andrezito [222]

There would be 8 green ones

5 0

4 years ago

Read 2 more answers

The manager of a pizza chain in Albuquerque, New Mexico, wants to determine the average size of their advertised 16-inch pizzas.

erastova [34]

Answer:

n=50

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".

$\bar X=16.10$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=1.8 represent the sample standard deviation

n=25 represent the sample size

2) 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)

We need to find the degrees of freedom given by:

$df=n-1=25-1=24$

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 $z_{\alpha/2}=\pm 1.96$

Since we assume that we are taking a bigger sample then we can replace the t distribution with the normal standard distribution, and we can assume that th population deviation is 1.8. 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 =0.5 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)

Replacing into formula (b) we got:

$n=(\frac{1.96(1.8)}{0.5})^2 =49.79 \approx 50$

So the answer for this case would be n=50 rounded up to the nearest integer

5 0

3 years ago