Ask question

Ask question

All categories

2 years ago

12

There are 16 books on a shelf. 7 of these books are new. (a) What is the ratio of new books to used books? (b) What is the ratio

of all books to used books?

2 answers:

777dan777 [17]2 years ago

4 0

The answers are
a. 7:9
b. 16:7

ELEN [110]2 years ago

4 0

7 out of the 16 books are new, so 9 of them are used.

7:9 is the ratio of new books to old books.
16:9 is the ratio of all books to used books.

You might be interested in

If if the branding division were to experience the same percentage increase from year 2 to year 3 as they did from year 1 to yea

Sedaia [141]

Since the basis is from year 1 to year 2, calculate first for the difference of their percentages. That would be:

Difference = year 2 - year 1
Difference = 2.32% - 1.1% = 1.22%

We apply this same value of percentage increase from year 2 to year. Thus, the percentage for year 3 is:

% Year 3 = % Year 2 + percentage increase
% Year 3 = 2.32% + 1.22%
% Year 3 = 3.54%

6 0

2 years ago

Question 1 (1 point) Point M is the midpoint of AB. The coordinates of point A are (-6, 3) and the coordinates of M are (-1, 1).

konstantin123 [22]

Answer:

The answer to your question is: (4, -1)

Step-by-step explanation:

M = midpoint (-1, 1)

A = (-6, 3)

B = (x, y)

$xm = \frac{x1 + x }{2}$

x = 2xm - x1

x = 2(-1) + 6

x = -2 + 6

x = 4

$ym = \frac{y1 + y }{2}$

y = 2ym - y1

y = 2(1) - 3

y = 2 - 3

y = -1

4 0

3 years ago

A company's stock was selling at

skad [1K]

Answer:

20% increase

Step-by-step explanation:

30-25=5

$\frac{5}{25}=\frac{x}{100}$

solve for x

(5*100)/25 = 20

8 0

3 years ago

Question1, shop A normal price £85 1/3 off and shop B normal price £80 30% off

ser-zykov [4K]

#1 shop A would be cheaper

6 0

3 years ago

PLEASE HELP! The table shows the number of championships won by the baseball and softball leagues of three youth baseball divisi

Irina18 [472]

Answer:

Question 1: $P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16}}{ \frac{4}{16}} = \frac{1}{2}$

Question 2:

A. $P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}$

B. $P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}$

C. $P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}$

Step-by-step explanation:

Conditional probability is defined by

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

with P(A and B) beeing the probability of both events occurring simultaneously.

Question 1:

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

$P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16} }{ \frac{4}{16} } = \frac{1}{2}$

Question 2.A:

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

$P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}$

Question 2.B:

Z: Championships won by the 13 - 15 years old, beeing

$P ( Z)= \frac{ 1 }{ 16 }$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

then

P( Z and B)= \frac{ 1 }{ 16 }[/tex]

By definition,

$P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}$

Question 3.B

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

Z: Championships won by the 13 - 15 years old, beeing

$P ( Z)= \frac{ 1 }{ 16 }$

then

$P (Y or Z) = P(Y) + P(Z) = \frac{6}{16}$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

so

$P((YorZ) and B)= \frac{3}{16}$

By definition,

$P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}$

3 0

3 years ago