Ask question

Ask question

All categories

2 years ago

6

Mrs. Wood is a librarian at Westside Library. In examining a random sample of the library's book collection, she found the follo

wing. 716 books had no damage, 68 books had minor damage, and 43 books had major damage. Based on this sample, how many of the 36,500 books in the collection should Mrs. Wood expect to have major damage? Round your answer to the nearest whole number. Do not round any intermediate calculations.

1 answer:

uysha [10]2 years ago

6 0

Using proportions, it is found that Mrs. Wood should expect 1,898 books out of the 36,500 to have major damage.

<h3>What is a proportion?</h3>

A proportion is a fraction of a total amount, and the measures are related using a rule of three.

For this problem, the proportion of books with major damage is:

43/(716 + 68 + 43) = 0.051995.

Hence, out of 36,500 books, the expected amount of books with major damage is:

0.051995 x 36,500 = 1898.

More can be learned about proportions at brainly.com/question/24372153

#SPJ1

You might be interested in

If JKL= MNP, which statement must be true

Aleks04 [339]

B is correct answer.

8 0

2 years ago

Read 2 more answers

Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times t

Anika [276]

Answer:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

Step-by-step explanation:

Given

$\int\limits {\frac{x}{x^4 + 16}} \, dx$

Required

Solve

Let

$u = \frac{x^2}{4}$

Differentiate

$du = 2 * \frac{x^{2-1}}{4}\ dx$

$du = 2 * \frac{x}{4}\ dx$

$du = \frac{x}{2}\ dx$

Make dx the subject

$dx = \frac{2}{x}\ du$

The given integral becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{x}{x^4 + 16}} \, * \frac{2}{x}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{1}{x^4 + 16}} \, * \frac{2}{1}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

Recall that: $u = \frac{x^2}{4}$

Make $x^2$ the subject

$x^2= 4u$

Square both sides

$x^4= (4u)^2$

$x^4= 16u^2$

Substitute $16u^2$ for $x^4$ in $\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16u^2 + 16}} \,\ du$

Simplify

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{2}{16}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

In standard integration

$\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)$

So, the expression becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)$

Recall that: $u = \frac{x^2}{4}$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

4 0

2 years ago

Three scatterplots are shown, along with their best-fit lines and correlation coefficients, r1, r2, and r3. Choose the correct s

ira [324]

Answer:

C) r3 < r2 < r1

Step-by-step explanation:

8 0

3 years ago

In 8 minutes,josh was able to type 496 words.What's his typing speed in words per minute

leva [86]

62 words per minute. You would divide 8 from 496 since you need to find out how many words he can type in a minute. Your answer is 62 words per minute.

Hope this helped! :)

3 0

3 years ago

Use the Midpoint Rule and the given data to estimate the value of the integral I. (Give the answer to two decimal places.)

svlad2 [7]

Answer:

a) integral = 24.72

b) |Error| ≤ 0.4267

Step-by-step explanation:

a)

The integral:

$\int_{0}^{3.2} f(x) dx$

can be approximated with the midpoint rule, as follows:

6.7*(0.8 - 0.0) + 8.9*(1.6 - 0.8) + 6.9*(2.4-1.6) + 8.4*(3.2 - 2.4) = 24.72

(that is, all the intervals are 0.8 units length and f(x) is evaluated in the midpoint of the interval)

b)The error bound for the midpoint rule with <em>n</em> points is:

|Error| ≤ K*(b - a)^3/(24*n^2)

where <em>b</em> and are the limits of integration of the integral and K = max |f''(x)|

Given that -5 ≤ f''(x) ≤ 1, then K = 5. Replacing into the equation:

|Error| ≤ 5*(3.2 - 0)^3/(24*4^2) = 0.4267

5 0

2 years ago