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2 years ago

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Y = 2x + 3 2 How does the equation of a line need to be changed to shift the line upward? 1 123 1 -3 -2 -1 -1 -2 A A. The absolu

te value of the slope needs to be increased. B. The absolute value of the slope needs to be decreased. C. The y-intercept needs to be increased.

1 answer:

Jobisdone [24]2 years ago

8 0

Answer:

C

Step-by-step explanation:

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A box of cookies cost $2 before tax. Find the after-tax cost if the sales tax was 8.5%.

adell [148]

Answer:

$2.17 total

Step-by-step explanation:

2 (0.85) = v2.17

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3 years ago

What is the domain of the cubic function -(2x+3)^3 -1

Mazyrski [523]

Answer:

(-infinite,+infinite)

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3 years ago

At Munder Difflin Paper Company, the manager Mitchell Short randomly places golden sheets of paper inside of 30% of their paper

Korvikt [17]

Answer:

90.67% probability that John finds less than 7 golden sheets of paper

Step-by-step explanation:

For each container, there are only two possible outcomes. Either it contains a golden sheet of paper, or it does not. The probability of a container containing a golden sheet of paper is independent of other containers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

At Munder Difflin Paper Company, the manager Mitchell Short randomly places golden sheets of paper inside of 30% of their paper containers.

This means that $p = 0.3$

14 of these containers of paper.

This means that $n = 14$

What is the probability that John finds less than 7 golden sheets of paper?

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{14,0}.(0.3)^{0}.(0.7)^{14} = 0.0068$

$P(X = 1) = C_{14,1}.(0.3)^{1}.(0.7)^{13} = 0.0407$

$P(X = 2) = C_{14,2}.(0.3)^{2}.(0.7)^{12} = 0.1134$

$P(X = 3) = C_{14,3}.(0.3)^{3}.(0.7)^{11} = 0.1943$

$P(X = 4) = C_{14,4}.(0.3)^{4}.(0.7)^{10} = 0.2290$

$P(X = 5) = C_{14,5}.(0.3)^{5}.(0.7)^{9} = 0.1963$

$P(X = 6) = C_{14,6}.(0.3)^{6}.(0.7)^{8} = 0.1262$

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0068 + 0.0407 + 0.1134 + 0.1943 + 0.2290 + 0.1963 + 0.1262 = 0.9067$

90.67% probability that John finds less than 7 golden sheets of paper

7 0

2 years ago

HELP PLEASE!!!! 8TH GRADE MATH

vladimir2022 [97]

1. 12 pages per minute

3 0

3 years ago

Tres compañías de navegación pasan por cierto puerto. La primera cada 8 días; la

lilavasa [31]

I think this is a least common multiple question

So they would all arrive on day 504, (which probably is a few years)

8 = 2 × 2 × 2

18 = 2 × 3 × 3

21 = 3 × 7

I might be right I might be wrong, IM not positively sure but yknow

4 0

2 years ago