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Serhud [2]
3 years ago
10

GODD SOMEONE HELP ME RIGHT NOW PLEASE THIS IS DUE TODAY AND I AM DUM I GIVE POINTS AND BRAINLIEST JUST HELP MEEE

Mathematics
1 answer:
Sedbober [7]3 years ago
4 0

Answer:

-5 and -6

Step-by-step explanation:

For trinomial factoring, you need two numbers that add up to get the -11x and two numbers that multiply to get 30.

-5 + (-6) = -11

-5 × -6 = 30

You might be interested in
If 3x+54+(65 × 32)+ 902- (43/12) =y, then what is y?
IceJOKER [234]

Answer:

We cannot solve without a value for x. The simplified form is 3x+\frac{36,389}{12} =y

Step-by-step explanation:

If the equation is 3x+54+(65)(32)+902-\frac{43}{12}=y then we cannot solve it until we know the value of x. We can simplify it using order of operations such as PEMDAS.

3x+54+(65)(32)+902-\frac{43}{12}=y \\3x+54+2,080+902-\frac{43}{12}=y \\3x+3,036-\frac{43}{12}=y \\3x+\frac{36,432}{12} -\frac{43}{12} =y\\3x+\frac{36,389}{12} =y

6 0
3 years ago
Return to the credit card scenario of Exercise 12 (Section 2.2), and let C be the event that the selected student has an America
Nadya [2.5K]

Answer:

A. P = 0.73

B. P(A∩B∩C') = 0.22

C. P(B/A) = 0.5

   P(A/B) = 0.75

D. P(A∩B/C) = 0.4

E. P(A∪B/C) = 0.85

Step-by-step explanation:

Let's call A the event that a student has a Visa card, B the event that a student has a MasterCard and C the event that a student has a American Express card. Additionally, let's call A' the event that a student hasn't a Visa card, B' the event that a student hasn't a MasterCard and C the event that a student hasn't a American Express card.

Then, with the given probabilities we can find the following probabilities:

P(A∩B∩C') = P(A∩B) - P(A∩B∩C) = 0.3 - 0.08 = 0.22

Where P(A∩B∩C') is the probability that a student has a Visa card and a Master Card but doesn't have a American Express, P(A∩B) is the probability that a student has a has a Visa card and a MasterCard and P(A∩B∩C) is the probability that a student has a Visa card, a MasterCard and a American Express card. At the same way, we can find:

P(A∩C∩B') = P(A∩C) - P(A∩B∩C) = 0.15 - 0.08 = 0.07

P(B∩C∩A') = P(B∩C) - P(A∩B∩C) = 0.1 - 0.08 = 0.02

P(A∩B'∩C') = P(A) - P(A∩B∩C') - P(A∩C∩B') - P(A∩B∩C)

                   = 0.6 - 0.22 - 0.07 - 0.08 = 0.23

P(B∩A'∩C') = P(B) - P(A∩B∩C') - P(B∩C∩A') - P(A∩B∩C)

                   = 0.4 - 0.22 - 0.02 - 0.08 = 0.08

P(C∩A'∩A') = P(C) - P(A∩C∩B') - P(B∩C∩A') - P(A∩B∩C)

                   = 0.2 - 0.07 - 0.02 - 0.08 = 0.03

A. the probability that the selected student has at least one of the three types of cards is calculated as:

P = P(A∩B∩C) + P(A∩B∩C') + P(A∩C∩B') + P(B∩C∩A') + P(A∩B'∩C') +              

     P(B∩A'∩C') + P(C∩A'∩A')

P = 0.08 + 0.22 + 0.07 + 0.02 + 0.23 + 0.08 + 0.03 = 0.73

B. The probability that the selected student has both a Visa card and a MasterCard but not an American Express card can be written as P(A∩B∩C') and it is equal to 0.22

C. P(B/A) is the probability that a student has a MasterCard given that he has a Visa Card. it is calculated as:

P(B/A) = P(A∩B)/P(A)

So, replacing values, we get:

P(B/A) = 0.3/0.6 = 0.5

At the same way, P(A/B) is the probability that a  student has a Visa Card given that he has a MasterCard. it is calculated as:

P(A/B) = P(A∩B)/P(B) = 0.3/0.4 = 0.75

D. If a selected student has an American Express card, the probability that she or he also has both a Visa card and a MasterCard is  written as P(A∩B/C), so it is calculated as:

P(A∩B/C) = P(A∩B∩C)/P(C) = 0.08/0.2 = 0.4

E. If a the selected student has an American Express card, the probability that she or he has at least one of the other two types of cards is written as P(A∪B/C) and it is calculated as:

P(A∪B/C) = P(A∪B∩C)/P(C)

Where P(A∪B∩C) = P(A∩B∩C)+P(B∩C∩A')+P(A∩C∩B')

So, P(A∪B∩C) = 0.08 + 0.07 + 0.02 = 0.17

Finally, P(A∪B/C) is:

P(A∪B/C) = 0.17/0.2 =0.85

4 0
3 years ago
This figure represents a specialty design for an art print, made of a square and 3 identical triangles.
Gnoma [55]

Answer:

91cm2 (or just 91)

Step-by-step explanation:

the area of the square is 49, and 1 triangle is 14 you have to add all 3 triangles plus the square(or the base).

8 0
2 years ago
Read 2 more answers
Need help with 13 14 and 15
ad-work [718]

Answer:

njbvi3brfjkncdoewkmdrfrj

Step-by-step explanation:

gagagagagaggag

6 0
3 years ago
A college job placement office collected data about students’ GPAs and the salaries they earned in their first jobs after gradua
frez [133]

Answer:

X is the GPA

Y is the Salary

Standard deviation of X is 0.4

Standard deviation of Y is 8500

E(X)=2.9

E(Y)=47200

We are given that The correlation between the two variables was r = 0.72

a)y = a+bx

b = \frac{\sum(x_i-\bar{x})(y_i-\bar{y})}{\sum(x_i-\bar{x})^2} = \frac{r \times \sqrt{var(X) \times Var(Y)}}{Var(X)} =  \frac{0.72 \times \sqrt{0.4^2 \times 8500^2}}{0.4^2} = 15300

a=y-bx = 47200-(15300 \times 29) = 2830

So, slope =  15300

Intercept =  2830

So, equation : y = 2830+15300x

b) Your brother just graduated from that college with a GPA of 3.30. He tells you that based on this model the residual for his pay is -$1880. What salary is he earning?

y = 2830+15300 \times 3.3 = 53320

Observed salary = Residual + predicted = -1860+53320 = 51440

c)) What proportion of the variation in salaries is explained by variation in GPA?

The proportion of the variation in salaries is explained by variation in GPA = r^2 = (0.72)^2 =0.5184

8 0
3 years ago
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