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3 years ago

Julia charged $612 to her credit card. The cost of her next purches caused her to go over her credit limit of$2000.

Mathematics

1 answer:

trasher [3.6K]3 years ago

6 0

612+C=2000 then just subtract 612 from 2000 and get the value of C.

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Substitute the values for a, b, and c into b2 – 4ac to determine the discriminant. Which quadratic equations will have two real

garik1379 [7]

The complete question is

"Substitute the values for a, b, and c into b2 – 4ac to determine the discriminant. Which quadratic equations will have two real number solutions? (The related quadratic function will have two x-intercepts.) Check all that apply.

0 = 2x^2 – 7x – 9

0 = 4x^ 2 – 3x – 1

The quadratic equations that have real number solutions are; 4x^2 – 3x – 1, and 2x^2 – 7x – 9.

<h3>What is the formula for Discriminant?</h3>

The formula for finding the discriminant is

$b^2 - 4ac$

The solution contains the term $\sqrt{b^2 - 4ac}$ which will be:

Real and distinct if the discriminant is positive

Real and equal if the discriminant is 0

Non-real and distinct roots if the discriminant is negative

For the quadratic equation $2x^2 - 7x - 9$

$b^2 - 4ac$

$= (-7) ^2 - 4( 2) ( -9)\\\\= 49 + 72 = 121$

This equation has two real number solutions.

For the quadratic equation $4x^ 2 - 3x- 1$

$b^2 - 4ac$

$= (-3) ^2 - 4( 4) ( -1)\\\\= 9 + 16 = 25$

This equation will have two real number solutions.

Learn more on discriminant here:

brainly.com/question/1537997

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2 years ago

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( -1) 3^3

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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.

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

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

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