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Karolina [17]
2 years ago
13

Please help me w/ iready

Mathematics
1 answer:
kirill [66]2 years ago
8 0

Answer:

45 = 45

or

just 45.

Step-by-step explanation:

a = 6 , b = 2

3(6) + 4(2)^3 - 5 = 3(6) + 4(2)^3 - 5 (Given)

18 + 32 - 5 = 18 + 32 - 5 (Solve parenthesis)

45 = 45 (combine like terms)

45 does equal 45 so there is the solution. There is infinite many solutions.

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Which table represents only values described by the relationship between a and b?
algol13

Answer:

a

Step-by-step explanation:

8 0
3 years ago
3n + 9 = 27<br><br> n = ?<br><br> solve for n
Debora [2.8K]

Answer:

6

Step-by-step explanation:

7 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
A bus travels 36 miles in 45 minutes. Enter the number of miles the bus travels in 60 minutes at this rate.
kompoz [17]

Answer:

The answer would be 48 miles in 60 minutes. I hope that helps!

Step-by-step explanation:

36 x 60 = 2160

45n = 2160

n = 48

8 0
3 years ago
Read 2 more answers
Find the equation of locus of a point which moves such that its distance from (0,2) is one third distance from (-2,3). ( I WILL
scoray [572]

Answer:

8(x^2+y^2)-4x-30y+23=0

Step-by-step explanation:

<u />

<u>Distance formula</u>

\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Let P(x, y) = any point on the locus

Let A = (0, 2)    

Let B = (-2, 3)

If a point moves such that its distance from (0, 2) is one third distance from (-2, 3):

PA=\dfrac{1}{3}PB

Therefore, using the distance formula:

\implies \sqrt{(x-0)^2+(y-2)^2}=\dfrac{1}{3}\sqrt{(x-(-2))^2+(y-3)^2}

Square both sides:

\implies x^2+(y-2)^2=\dfrac{1}{9}[(x+2)^2+(y-3)^2]

\implies x^2+y^2-4y+4=\dfrac{1}{9}(x^2+4x+4+y^2-6y+9)

Multiply both sides by 9:

\implies 9x^2+9y^2-36y+36=x^2+4x+4+y^2-6y+9

\implies 8x^2+8y^2-4x-30y+23=0

\implies 8(x^2+y^2)-4x-30y+23=0

3 0
2 years ago
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