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Umnica [9.8K]
3 years ago
6

When Mr. Key made a chocolate milkshake for himself, he used 3/4 cup

Mathematics
1 answer:
icang [17]3 years ago
8 0

Answer:

3 cups of milk

Step-by-step explanation:

4/4 is 1 cup.

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-3x^2+6x+1=4 find the discriminant of each quadratic equation then state the number and type solutions
ipn [44]

Answer:

S={( 1 )}

Step-by-step explanation:

-3x²+6x+1=4

-3x²+6x+1-4=0

-3x²+6x-3=0

∆=b²-4.a.c

∆=(6)²-4.(-3).(-3)

∆=36-36

∆=0

x'=x"=-b/2a=-(+6)/2.(-3)=-6/-6=1

3 0
3 years ago
Read 2 more answers
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
Help Help Help Help...What does it take for an orbiting spacecraft to escape its orbit?
Nana76 [90]

Answer:

Manual Override??

Step-by-step explanation: I'm prop wrong

8 0
3 years ago
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How can you find this distance (-3,7),(0,4) in words
Bezzdna [24]

Answer:

The distance between these two given points is:

3\sqrt{2}

Step-by-step explanation:

We are given two points:

(-3,7),(0,4)

<em>The distance between two points (a,b) and (c,d) is given by the distance formula as:</em>

<em>\sqrt{(c-a)^2+(d-b)^2</em>

similarly we can find the length of a line segment by considering the distance between the end points of the line segment.

So here (a,b)=(-3,7)

and (c,d)=(0,4).

Hence distance between these two points is given by:

\sqrt{(0-(-3))^2+(4-7)^2}=\sqrt{(3)^2+(-3)^2}=\sqrt{9+9}\\   \\=\sqrt{18} \\\\=3\sqrt{2}

6 0
3 years ago
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I NEED REAL HELP QUICKLY please
skad [1K]

Answer:

hold on its its.............oh yeah 21

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