Turn the following fraction into a PERCENT: 15/60

The pie chart below shows the percentage of total revenue that a publisher receives from various types of publications. Use this

liberstina [14]

Answer:

a. Cookbooks

b. 50%

c. 30%

Step-by-step explanation:

According To the Question,

We have a total revenue of a publisher describe in a circle (360°).

a. Now, Approximately Cookbooks & Textbooks revenue Form 180°, but textbook revenue is more than cookbooks as clearly visible in the diagram

Thus, cookbooks are less than 90°.

Now, we have to find 1/5th of total publisher revenue which is 360°/5= 72°. & the Cookbooks is nearest to 72° ( less than 90°)

So, Answer For (a) is Cookbooks

b. Here in Diagram Clearly Visible that The Cookbooks & Textbooks Revenue Form 180° which is Approximately 50% of total Publisher's Revenue.

So, the Total Revenue Comes From Textbooks & Cookbooks is 50%.

c. Now We know the Cookbooks + Textbooks Revenue form 180° Approximately & Cookbook is approximately equal to 72° (as we solve above)

So, textbooks are 180°-72° = 108°, which is 30% of 360° ∴ 30% Revenue Come From Textbooks.

7 0

2 years ago

The Venn diagram shows the results of two events resulting from rolling a number cube.

tankabanditka [31]

Answer:

$P(A|B)=\frac{2}{3}$

$P(A)*P(B)=\frac{1}{3}$

$P(A) =\frac{2}{3}$

$P(B) =\frac{1}{2}$ .

Step-by-step explanation:

We use the Venn diagram to calculate the desired probabilities.

Note that there are 6 possible results in the sample space

S = {1, 2, 3, 4, 5, 6}

Then note that in the region representing the intercept of A and B there are two possible values.

So

$P (A\ and\ B) = \frac{2}{6} = \frac{1}{3}$

In the region that represents event A there are 4 possible outcomes {4, 5, 1, 2}

So

$P(A) = \frac{4}{6} = \frac{2}{3}$

In the region that represents event B there are 3 possible outcomes {1, 2, 6}

So

$P(B) = \frac{3}{6} = \frac{1}{2}$ .

Now

$P(A | B)=\frac{P(A \ and\ B)}{P(B)}\\\\P(A | B)=\frac{\frac{1}{3}}{\frac{1}{2}}\\\\P(A|B)=\frac{2}{3}$

$P(A)*P(B)=\frac{2}{3}*\frac{1}{2}=\frac{1}{3}$

6 0

2 years ago

Becky is an experienced swimmer. The table contains data on her times in the 100-meter freestyle event for each year over the pa

Sauron [17]

Correlation coefficient (r) = [nΣxy - (Σx)(Σy)] / [sqrt(nΣx^2 - (Σx)^2)sqrt(nΣy^2 - (Σy)^2)]
Σx = 21 => (Σx)^2 = 21^2 = 441
Σy = 671 => (Σy)^2 = 671^2 = 450,241
Σx^2 = 1 + 4 + 9 + 16 + 25 + 36 = 91
Σy^2 = 98^2 + 101^2 + 109^2 + 117^2 + 119^2 + 127^2 = 75,665
Σxy = 1(98) + 2(101) + 3(109) + 4(117) + 5(119) + 6(127) = 2,452
r = [6(2,452) - 21(671)] / [sqrt(6(91) - 441)sqrt(6(75,665) - 450,241)] = 621/sqrt(105)sqrt(3749) = 0.99

option b is the correct answer.

5 0

2 years ago

A candy manufacturer makes two types of special candy, say A and B. Candy A consists of equal parts of dark chocolate and carame

Sphinxa [80]

If we let x as candy A
y as candy B
a as dark chocolate in candy a
b as dark chocolate in candy b
c as caramel
d as walnut
P as profit
we have the equations:
a + c = x
2b + d = y
a + 2b ≤ 360
c ≤ 430
d ≤ 210
P = 285x + 260y

This is an optimization problem which involves linear programming. It can be solved by graphical method or by algebraic solution.
P = 285(a + c) + 260(2b +d)
If we assume a = b
Then a = 120, 2b = 240
P = 285(120 + 120) + 260(240 + 120)
P = 162000
candy A should be = 240
candy B should be = 360

6 0

2 years ago

The first three terms of an arithmetic series are 6p+2, 4p²-10 and 4p+3 respectively. Find the possible values of p. Calculate t

Doss [256]

Answer:

First Case:

$\displaystyle p=\frac{5}{2}\text{ and } d=-2$