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3 years ago

2x² + 9 = - 7x how do i find an answer using the quadratic formula

Mathematics

1 answer:

jok3333 [9.3K]3 years ago

7 0

Answer:

(-7 ± $i$ √(-23)) / 4

Step-by-step explanation:

Unless you have to show your work, you don't really have to use the quadratic formula unless it is unfactorable.

2x² + 9 = -7x (Add 7x to both sides)

2x² + 7x + 9 = 0 ( The quadratic formula is (-b ± √(b^2 - 4ac))/2a )

A = 2

B = 7

C = 9

(-7 ± √(49 - 72))/4

= (-7 ± √(-23))/4 (Which simplifies to an imaginary answer, proving that this is, indeed, unfactorable)

= (-7 ± $i$ √(-23))/4

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2 years ago

A recent college graduate is planning to take the first three actuarial examinations in the coming summer. She will take the fir

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(a) 0.7

(b) $\frac 13$

Step-by-step explanation:

Let $E_1, E_2,$ and $E_3$ be the events of passing the 1st, 2nd, and 3rd exam individually.

Given that $P(E_1)=0.9\;\cdots(i)$

where $P(E_1)$ denotes the probability of passing the 1st exam.

Condition for passing the second exam is, at first, the candidate must have to pass the 1st exam.

So, $P\left(\frac{E_2}{E_1}\right)=0.8\;\cdots(ii)$

where $P(E_2/E_1)$ denotes the probability of passing the 2nd exam when she already passed the 1st exam (given, 0.8).

Similarly, as the conditional probability of passing the 3rd exam is 0.7, and the condition for this is, at first, she must have to pass the 1st and 2nd exam. i.e,

$P\left(\frac{E_2}{P(E_2/E_1)}\right)=0.7\;\cdots(iii)$

(a) For passing all the exams, the condition is, at first, she has to pass the 1st and 2nd exam, then she has to pass the 3rd exam too. The probability for this conditional has been given as 0.7.

So, the probability that she passes all three exams is 0.7.

(b) Given that she didn't pass all three exams that means she either failed in 1st exam or she passed the 1st and failed in 2nd exam or she passed both 1st and 2nd but failed in the 3rd exam.

Let F be the event that she didn't pass all three exams. So,

$P(F)=(1-P(E_1))+\left(1-\frac{P(E_2)}{P(E_1)}\right)+\left(1-\frac{P(E_2)}{P(E_2/E_1)}}\right)$

$\Rightarrow P(F)=(1-0.9)+(1-0.8)+(1-0.7)=0.6$

Lef $F_2$ be the event that she failed the 2nd exam, so

$P(F_2)=1-\frac{P(E_2)}{P(E_1)}$

$\Rightarrow P(F_2)=1-0.8=0.2$

So, the conditional probability that she failed the 2nd exam is

$P\left(\frac{F_2}{F}\right)=\frac{P(F_2)}{P(F)}$

$\Rightarrow P\left(\frc{F_2}{F}\right)=\frac{0.2}{0.6}=\frac{1}{3}=0.33$

5 0

3 years ago

I don’t know if there is an outlier or not how can you tell

SIZIF [17.4K]

To find an outlier, multiply the IQR by 1.5

172,500 * 1.5 = 258,750

now, u add that to Q3.....545,000 + 258,750 = 803,750....anything above this number is an outlier

now,u subtract that from Q1....372,500 - 258,750 = 113,750....anything below that is an outlier

u have already done this...u have found the higher end and the lower end....so anyting larger then the higher end or lower then the lower end is ur outlier

so in ur data, there is only 1 outlier and that is 933.....130 is not an outlier because it is not less then 113,750.

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3 years ago

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Answer:

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Step-by-step explanation:

Recalling the standard deviation formula,

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Where sqrt = square root

n = Number of derivative

Therefore,:

if, H0: p is not 48% vs H1: p1 = 48%

standard deviation is:

σ = sqrt[ P * ( 1 - P ) / n ] = sqrt[ .48*.49/900 ] = .01666333+ z -score is (p-P)/σ

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ICONCLUSION:

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