What is a fraction that is more than 1/10 but less than 1/2

iVinArrow [24]

Answer:

5

Step-by-step explanation:

When squaring the number 5 (i.e $5^2$ ) the answer is 25. $5^{2}$ is the same thing as 5 times 5. Basically with the square root, you are going backwards, so you are therefore finding the base of it all.

7 0

3 years ago

Read 2 more answers

If

Nitella [24]

Since profit can't be negative, the production level that'll maximize profit is approximately equal to 220.

<h3>How to find the production level that'll maximize profit?</h3>

The cost function, C(x) is given by 12000 + 400x − 2.6x² + 0.004x³ while the demand function, P(x) is given by 1600 − 8x.

Next, we would differentiate the cost function, C(x) to derive the marginal cost:

C(x) = 12000 + 400x − 2.6x² + 0.004x³

C'(x) = 400 − 5.2x + 0.012x².

Also, revenue, R(x) = x × P(x)

Revenue, R(x) = x(1600 − 8x)

Revenue, R(x) = 1600x − 8x²

Next, we would differentiate the revenue function to derive the marginal revenue:

R'(x) = 1600 - 8x

At maximum profit, the marginal revenue is equal to the marginal cost:

1600 - 8x = 400 − 5.2x + 0.012x

1600 - 8x - 400 + 5.2x - 0.012x² = 0

1200 - 2.8x - 0.012x² = 0

0.012x² + 2.8x - 1200 = 0

Solving by using the quadratic equation, we have:

x = 220.40 or x = -453.73.

Since profit can't be negative, the production level that'll maximize profit is approximately equal to 220.

Read more on maximized profit here: brainly.com/question/13800671

#SPJ1

6 0

2 years ago

To estimate the mean height μ of male students on your campus,you will measure an SRS of students. You know from government data

nexus9112 [7]

Answer:

a) $\sigma = 0.167$

b) We need a sample of at least 282 young men.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

This Zscore is how many standard deviations the value of the measure X is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

(a) What standard deviation must x have so that 99.7% of allsamples give an x within one-half inch of μ?

To solve this problem, we use the 68-95-99.7 rule. This rule states that:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviations of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we want 99.7% of all samples give X within one-half inch of $\mu$ . So $X - \mu = 0.5$ must have $Z = 3$ and $X - \mu = -0.5$ must have $Z = -3$ .

So

$Z = \frac{X - \mu}{\sigma}$

$3 = \frac{0.5}{\sigma}$

$3\sigma = 0.5$

$\sigma = \frac{0.5}{3}$

$\sigma = 0.167$

(b) How large an SRS do you need to reduce the standard deviationof x to the value you found in part (a)?

You know from government data that heights of young men are approximately Normal with standard deviation about 2.8 inches. This means that $\sigma = 2.8$

The standard deviation of a sample of n young man is given by the following formula

$s = \frac{\sigma}{\sqrt{n}}$

We want to have $s = 0.167$

$0.167 = \frac{2.8}{\sqrt{n}}$

$0.167\sqrt{n} = 2.8$

$\sqrt{n} = \frac{2.8}{0.167}$

$\sqrt{n} = 16.77$

$\sqrt{n}^{2} = 16.77^{2}$

$n = 281.23$

We need a sample of at least 282 young men.

6 0

3 years ago

A group of students were surveyed to find out if they like building snowmen or skiing as a winter activity. The results of the s

soldier1979 [14.2K]

Building Snowmen - 60
Building Snowmen (no skiing) - 10
Skiing - 80
No to building snowmen - 50

Part A:
total of students: 200
140 x
----- = ----
200 100
14000 200x
--------- = --------
200 200
70 = x
70%

8 0

3 years ago

Read 2 more answers

Horn lengths of Texas longhorn cattle are normally distributed. The mean horn spread is 60 inches with a standard deviation of 4

german

Answer:

range is between 55.5 to 64.5

Step-by-step explanation:

Horn lengths of Texas longhorn cattle are normally distributed. The mean horn spread is 60 inches with a standard deviation of 4.5 inches

68% is 1 standard deviation from mean

To get the range of 1 standard deviation we add and subtract standard deviation from mean

mean = 60

standard deviation = 4.5

60 - 4.5= 55.5

60+4.5 = 64.5

1 standard deviation is between 55.5 to 64.5

That is 68% range is between 55.5 to 64.5

8 0

3 years ago