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2 years ago

Pls it 100 points Y= (X-3)(X+1)(X-2)

Mathematics

2 answers:

Strike441 [17]2 years ago

8 0

(x-3)(x+1)(x-2)
=(x-3)(x+1)(x+-2)
=(x-3)(x+1)(x)+(x-3)(x+1)(-2)
=x^3-2x^2-3x-2x^2+4x+6
=x^3-4x^2+x+6

nordsb [41]2 years ago

4 0

Answer:

x is 2 and y is 0

Step-by-step explanation:

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there are 16 tables in the lunch room.each tables can seat 22 studenrs. how many students can be seated at the at one time

masha68 [24]

16*22=352
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2 years ago

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Which of the following are exterior angles? check all that apply

Zigmanuir [339]

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4, 5, 6

Step-by-step explanation:

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3 years ago

A student answered 86 problems on a test correctly and received a grade of 98%. How many problems were on the test, if all the p

balu736 [363]

Answer:

88 problems

Step-by-step explanation:

Set this problem up as fractions

86/x = 98/100

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86 × 100 = x × 98

8600 = 98x

Divide both sides by 98

87.75... = x

Round to nearest whole number

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3 years ago

What Force would be needed to extend a spring with a spring constant k=7 N/m that extends<br>0.3m?

vladimir2022 [97]

Answer:

Force = - 2.1 Newton

Step-by-step explanation:

According to question ,

The value of Spring constant , K = 7 N/m Newton per meter

And the spring is extended up to 0.3 meter

The Force experience on the spring is calculated from the Hook's Law

<u>The principal state that , the force applied on spring to extend or compress the spring to some distance is proportion to that distance. </u>

Let the force be F

So, F = - K X

where K is spring constant and X is its displacement

Negative sign shows displacement of spring once it is stretched

And K is spring constant

So ,Force = ( - ) 7 × 0.3

I.e Force = - 2.1 Newton Answer

4 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