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lutik1710 [3]
3 years ago
10

How to solve 5(x+2)(x-2)

Mathematics
1 answer:
nevsk [136]3 years ago
7 0
The answer is: 5x^2-20
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32, percentage increaced by 25% =40

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X≥1 x≤7 y≥-1/3x+6
Goshia [24]

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if you were to set the boundaries of the feasibility region it would be as follows:

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Clarisse used the ordered pairs (0, 13) and (24, 5) for her trend line. What would be the y-intercept of her trend line?
laila [671]

Answer:

Step-by-step explanation:

Remark

You have 2 points to solve the equation y = mx + b. After finding m, use either of the points to find b.

Formula

y = mx + b

m = (y2 - y1)/(x2 - x1)

Solution

  • y2 = 13
  • y1 = 5
  • x2 = 0
  • x1 = 24

m = (13 - 5)/(0 - 24)

m = 8 / - 24

m = - 1/3

y = mx + b

y = -1/3 x + b

Use (0,13)  as the point.

13 = -1/3 * 0 + b

13 = b

The y intercept = (0,13)

5 0
2 years ago
A marketing firm is considering making up to three new hires. Given its specific needs, the management feels that there is a 50%
IRISSAK [1]

Answer:

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)

P(x\geq 2) = P(x=2) + P(x=3)\\0.5 = P(x=2) + 0.18\\ P(x=2) = 0.32

Also,

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

7 0
3 years ago
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