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meriva
3 years ago
11

Which is the direct linear variation equation for the relationship?

Mathematics
1 answer:
blondinia [14]3 years ago
7 0
Thank you for posting your question here at brainly. Feel free to ask more questions.  
<span>The best and most correct answer among the choices provided by the question is C. y = 5x</span> .         <span>
<span>
Hope my answer would be a great help for you. </span> </span> <span> </span>
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(4*10^2)(2*10^5) written in scientific notation
Akimi4 [234]
<span>(4*10^2)(2*10^5)
= 8 x 10^7

hope it helps</span>
8 0
3 years ago
Read 2 more answers
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

3 0
3 years ago
PLS HELP i have to finish this tonight
kvasek [131]

Answer:

D.480 Ft/S

Step-by-step explanation:

um, Just ask your teacher if you don't understand:P

5 0
3 years ago
Somone help me on please and thank you
Slav-nsk [51]

Answer:

You need to capitalize Europe

5 0
3 years ago
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Help me guys I suck at this :(
Lady_Fox [76]

Answer:

D.

Step-by-step explanation:

Breianliest?

7 0
3 years ago
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