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Lilit [14]
2 years ago
13

If 1/6 of the books in the library are cookbooks and 2/5 of the cookbooks are vegetarian cookbooks what fraction of the books in

the library are vegetarian cookbooks
A. 1/30 B. 1/15 C. 1/2 D. 3/10
Mathematics
2 answers:
Serga [27]2 years ago
8 0

Answer:

A. 1/30 of the cookbook

STALIN [3.7K]2 years ago
6 0

Answer:

1/15

Step-by-step explanation:

Took the quiz

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katen-ka-za [31]
I believe it would be 4.8 hours
5 0
1 year ago
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In a certain region, the number of highway accidents increased by 20% over the four years period. How many accidents were there
erma4kov [3.2K]

Equation:

a  =    5120  

r     =     1  +    .2     =     1 . 2

f (n )   =     5120 *  1 . 2^n - 1

N    =          4      years


f ( 4 )  5120 *  1 . 2^4 - 1

5120 *  1 . 2^3

We want to know, The number of Highway Accidents in Four Years.

Answer:    8,847 Highway Accidents total:






Hope that helps!!!!!!!!!!!                                                            : )


7 0
1 year ago
Kai has begun to list, in ascending order, the positive integers which are not factors of
Keith_Richards [23]

Answer:17

7, 9, 11, 13, 14, 17,

6 0
1 year ago
According to VSEPR theory, if there are five electron domains in the valence shell of an atom, they will be arranged in a(n) ___
andriy [413]

Answer:

trigonal bypyramidal

Step-by-step explanation:

It's difficult to explain without a drawing or a model, but it would basically arrange itself into two triangle based pyramids connected to each other at the base, which is trigonal bypyramidal.

5 0
1 year ago
1 point) As reported in "Runner's World" magazine, the times of the finishers in the New York City 10 km run are normally distri
BigorU [14]

Answer:

(a) E (X) = 61 and SD (X) = 9

(b) E (Z) = 0 and SD (Z) = 1

Step-by-step explanation:

The time of the finishers in the New York City 10 km run are normally distributed with a mean,<em>μ</em> = 61 minutes and a standard deviation, <em>σ</em> = 9 minutes.

(a)

The random variable <em>X</em> is defined as the finishing time for the finishers.

Then the expected value of <em>X</em> is:

<em>E </em>(<em>X</em>) = 61 minutes

The variance of the random variable <em>X</em> is:

<em>V</em> (<em>X</em>) = (9 minutes)²

Then the standard deviation of the random variable <em>X</em> is:

<em>SD</em> (<em>X</em>) = 9 minutes

(b)

The random variable <em>Z</em> is the standardized form of the random variable <em>X</em>.

It is defined as:Z=\frac{X-\mu}{\sigma}

Compute the expected value of <em>Z</em> as follows:

E(Z)=E[\frac{X-\mu}{\sigma}]\\=\frac{E(X)-\mu}{\sigma}\\=\frac{61-61}{9}\\=0

The mean of <em>Z</em> is 0.

Compute the variance of <em>Z</em> as follows:

V(Z)=V[\frac{X-\mu}{\sigma}]\\=\frac{V(X)+V(\mu)}{\sigma^{2}}\\=\frac{V(X)}{\sigma^{2}}\\=\frac{9^{2}}{9^{2}}\\=1

The variance of <em>Z</em> is 1.

So the standard deviation is 1.

8 0
2 years ago
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