Find the quotient. -10d 4e3f 5 ÷ 15d 2e 2f

Which expdession is equivalent to 4-2(3x-5)

GalinKa [24]

Answer:

4

Step-by-step explanation:

Your answer would be 0 because you have to subtract 3 by 5 you get the answer 2. Then you subtract 4 by 2 and you 2. After that you multiply 2 by 2 getting the answer 4.

7 0

3 years ago

Find the value of -(-2x+1)=9-14x

Elodia [21]

2x - 1 = 9 - 14x

16x - 1= 9

16x = 10

x = 10/16
x= 5/8 is the answer

7 0

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

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

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

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

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

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

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

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

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

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