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11Alexandr11 [23.1K]
2 years ago
13

Which point is not on the graph of the equation y=10+x

Mathematics
1 answer:
RSB [31]2 years ago
6 0

Answer:

C. (8, 2)

Step-by-step explanation:

To find which point is not on the graph, find which set of coordinates makes the equation incorrect.

When plugged into the equation, the point (8, 2) makes it incorrect:

y = 10 + x

2 = 10 + 8

2 = 18

2 \neq 18

The equation is incorrect because 2 is not equal to 18.

So, the point that is not on the graph is C. (8, 2)

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In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it
gavmur [86]

Answer:

a) P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b) P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c)  A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

Step-by-step explanation:

Assuming the following table:

                                                     Purchased Gum      Kept the Money   Total

Students Given 4 Quarters              25                              14                      39

Students Given $1 Bill                       15                               29                    44

Total                                                   40                              43                     83

a. find the probability of randomly selecting a student who spent the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student spent the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{40}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{15}{83}

And if we replace we got:

P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b. find the probability of randomly selecting a student who kept the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student kept the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{43}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{29}{83}

And if we replace we got:

P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c. what do the preceding results suggest?

For this case the best solution is:

A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

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3 years ago
Canadian provinces by population density
Savatey [412]
Not quite sure what the question being asked here is.
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3 years ago
Does the fraction 1/2 make the equation +6/6=3/2 true?
dolphi86 [110]

Answer:

Yes

Step-by-step explanation:

1/2 + 6/6

= (1*3)/(2*3) + 6/6

= 3/6 + 6/6

= 9/6

= 3/2

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7 0
2 years ago
Read 2 more answers
Select the best equation needed to solve for x; then solve for x:
Natali [406]

Step-by-step explanation:

A. 5x=12x-14

you will  simplify the equation to the form, which is simple to understand  

5x=12x-14

We move all terms containing x to the left and all other terms to the right.  

+5x-12x=-14

We simplify left and right side of the equation.  

-7x=-14

We divide both sides of the equation by -7 to get x.  

x=2

B.   5B+12=BB=28

you will  move all terms to the left:

.5B+12-(BB)=0

you will  move all terms containing B to the left, all other terms to the right

.5B-BB=-12

7 0
3 years ago
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Raj gets 1.5 mile head start and runs at a rate of 4.5 miles per hour.Jacinda's progress is represented by a graph that goes thr
Anuta_ua [19.1K]
Jacinda:  d = 10 * t  ( a rate: 10 miles per hour )
Raj :  d = 1.5 + 4.5 * t
Jacinda will catch up with Raj when:
10 t = 1.5 + 4.5 t
10 t - 4.5 t = 1.5
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t = 1.5 : 5.5 = 0.27 hours = 16.2 min  =  16 min 12 seconds 
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