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Nitella [24]
3 years ago
8

2. At a college bookstore, Carla purchased a math textbook and a novel that cost a total of $54, not

Mathematics
1 answer:
Svetradugi [14.3K]3 years ago
3 0

Answer:

the price of the novel : $10.25

The price of the math book : $43.75

Step-by-step explanation:

Let M represent math book and N novel

the price of the math textbook, is $8 more than 3 times the price of the novel.

M + N = $54 and M = 3N + 8

we can write 3N + 8 instead of M

N + 3N + 8 = $54 add the like terms

4N = $46

N = $10.25 this is the price of the novel and the price of the math book is $54 - $10.25 = $43.75

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zalisa [80]

You sometimes need to rewrite a factor because the lead number is not between 1 and 10 as it needs to be in scientific notation. For instance, if we had lead numbers of 2 and 5, as we do in the following problem.

2x10^3 * 5x10^9

Then we get the answer of 10x10^12. We need to adjust this by moving the decimal place and increasing the exponent by 1. The correct answer would be 1x10^13 instead.

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3 years ago
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yaroslaw [1]

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I believe it would be - (n+4) x 2

Step-by-step explanation:

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A total of 8644 people went to the football game Of those people 5100sat on the home side and the rest sat on the visitors side
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Calculate the number of people who sat on the visitors' side:

 8644
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3544 people
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3 years ago
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A {(2,7), (2,8), (3,8)}
Aleksandr [31]

Answer:

b

Step-by-step explanation:

4 0
3 years ago
The prior probabilities for events A1 and A2 are P(A1) = 0.35 and P(A2) = 0.50. It is also known that P(A1 ∩ A2) = 0. Suppose P(
forsale [732]

Answer:

Step-by-step explanation:

Hello!

Given the probabilities:

P(A₁)= 0.35

P(A₂)= 0.50

P(A₁∩A₂)= 0

P(BIA₁)= 0.20

P(BIA₂)= 0.05

a)

Two events are mutually exclusive when the occurrence of one of them prevents the occurrence of the other in one repetition of the trial, this means that both events cannot occur at the same time and therefore they'll intersection is void (and its probability zero)

Considering that P(A₁∩A₂)= 0, we can assume that both events are mutually exclusive.

b)

Considering that P(BIA)= \frac{P(AnB)}{P(A)} you can clear the intersection from the formula P(AnB)= P(B/A)*P(A) and apply it for the given events:

P(A_1nB)= P(B/A_1) * P(A_1)= 0.20*0.35= 0.07

P(A_2nB)= P(B/A_2)*P(A_2)= 0.05*0.50= 0.025

c)

The probability of "B" is marginal, to calculate it you have to add all intersections where it occurs:

P(B)= (A₁∩B) + P(A₂∩B)=  0.07 + 0.025= 0.095

d)

The Bayes' theorem states that:

P(Ai/B)= \frac{P(B/Ai)*P(A)}{P(B)}

Then:

P(A_1/B)= \frac{P(B/A_1)*P(A_1)}{P(B)}= \frac{0.20*0.35}{0.095}= 0.737 = 0.74

P(A_2/B)= \frac{P(B/A_2)*P(A_2)}{P(B)} = \frac{0.05*0.50}{0.095} = 0.26

I hope it helps!

5 0
3 years ago
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