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3 years ago

6

A photograph is 6 inches by 9 inches. A frame shop charges $0.25 per inch for a silver frame. How much would it cost to buy a si

lver frame for the photograph?

2 answers:

just olya [345]3 years ago

8 0

Perimeter = 2width + 2length

p=2(6)+2(9)
p=12+18
p=30

You need 30 inches of silver.

30x.25=7.50

The answer is $7.50.

Slav-nsk [51]3 years ago

7 0

The answer would be £7.50

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The cost to mail a first-class letter used to be $0.40 the first ounce. Each additional ounce costs $0.20. Fractions of an ounce

Hitman42 [59]

$1.00 because the first ounce is 0.40 then you have to times the other 3 ounce by 0.20 then u would have $1.00

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3 years ago

1. How many real solutions does the quadratic equation x?+9x= -8 have?

likoan [24]

Answer:

This quadratic equation has 2 solutions.

Step-by-step explanation:

I assume the '?' in your question is meant to be power 2 (²), or else it would not be a quadratic equation. You could write it using the superscript version of 2.

We can solve this equation by expressing it in the form: ax² + bx + c

x² + 9x= -8

x² + 9x + 8 = 0

Now if you know the discriminant, you can simply plug in your values of a, b, and c to see how many solutions there are.

In this case, you would not need the discriminant as there are whole-number factors and hence this can simply be factorised.

x² + 9x + 8 = 0

(x + 8)(x + 1) = 0

For this equation to be true (= 0), x can equal -8 OR -1.

Hence, this quadratic equation has 2 solutions.

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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,

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

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slega [8]

Answer:

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Answer:

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3 years ago