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3 years ago

Give a numerical example that shows these are not the same expressions. 2(n+5) and 2n+5

Mathematics

2 answers:

Nana76 [90]3 years ago

8 0

Answer: Any number that isn’t 1, 2, -1, or -2 I believe.

Step-by-step explanation: In the first expression, you have to distribute the 2, making it equal to 2n + 10, whereas the first is just 2n + 5.

For example if you were to plug the number 3 into this, you would have 2(3+5), which is 2(8) = 16. Versus the other side, 2(3)+5 = 11

Papessa [141]3 years ago

7 0

Answer:

The two are different expressions because when you distibute, you multiply all terms inside the parenthesis with the 2 in the first one: 2(n+5) = 2n + 10.

2n + 10 ≠ 2n + 5.

Step-by-step explanation:

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a) 0.9

b) Mean = 1.58

Standard Deviation = 0.89

Step-by-step explanation:

We are given the following in the question:

A marketing firm is considering making up to three new hires.

Let X be the variable describing the number of hiring in the company.

Thus, x can take values 0,1 ,2 and 3.

$P(x\geq 2) = 50\%= 0.5\\P(x = 0) = 10\% = 0.1\\P(x = 3) = 18\% = 0.18$

a) P(firm will make at least one hire)

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$P(x= 0) +P(x= 1) + P(x= 2) + P(x= 3) = 1\\ 0.1 + P(x= 1) + 0.32 + 0.18 = 1\\ P(x= 1) = 1- (0.1+0.32+0.18) = 0.4$

$\text{P(firm will make at least one hire)}\\= P(x\geq 1)\\=P(x=1) + P(x=2) + P(x=3)\\ = 0.4 + 0.32 + 0.18 = 0.9$

b) expected value and the standard deviation of the number of hires.

$E(X) = \displaystyle\sum x_iP(x_i)\\=0(0.1) + 1(0.4) + 2(0.32)+3(0.18) = 1.58$

$E(x^2) = \displaystyle\sum x_i^2P(x_i)\\=0(0.1) + 1(0.4) + 4(0.32) +9(0.18) = 3.3\\V(x) = E(x^2)-[E(x)]^2 = 3.3-(1.58)^2 = 0.80\\\text{Standard Deviation} = \sqrt{V(x)} = \sqrt{0.8036} = 0.89$

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