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GREYUIT [131]
3 years ago
13

The two plots below show the heights of some sixth graders and some seventh graders of a school:/Users/tylerleaird/Desktop/Scree

n Shot 2017-04-25 at 10.53.09 AM.png.The mean absolute deviation (MAD) for the first set of data is 1.2 and the MAD for the second set of data is 0.7. Approximately how many times the variability in the heights of the seventh graders is the variability in the heights of the sixth graders? (Round all values to the tenths place.)
0.5
1.2
1.7
3.4
Mathematics
1 answer:
fenix001 [56]3 years ago
7 0
0.5 will be rounded to 1.0
1.2 will be rounded to 1.0
1.7 will be rounded to 2.0
3.4 will be rounded to 3.0
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(a) Consider a class with 30 students. Compute the probability that at least two of them have their birthdays on the same day. (
Galina-37 [17]

Answer:

a.) 0.7063

b.) 23

Step-by-step explanation:

a.)

Let X be an event in which at least 2 students have same birthday

     Y be an event in which no student have same birthday.

Now,

P(X) + P(Y) = 1

⇒P(X) = 1 - P(Y)

as we know that,

Probability of no one has birthday on same day = P(Y)

⇒P(Y) = \frac{365!}{(365)^{n} (365-n)! }      where there are n people in a group

As given,

n = 30

⇒P(Y) = \frac{365!}{(365)^{30} (365-30)! } = \frac{365!}{(365)^{30} (335)! } = 0.2937

∴ we get

P(X) = 1 - 0.2937 = 0.7063

So,

The probability that at least two of them have their birthdays on the same day  =  0.7063

b.)

Given, P(X) > 0.5

As

P(X) + P(Y) = 1

⇒P(Y) ≤ 0.5

As

P(Y) = \frac{365!}{(365)^{n} (365-n)! }

We use hit and trial method

If n = 1 , then

P(Y) = \frac{365!}{(365)^{1} (365-1)! } = \frac{365!}{(365)^{1} (364)! }  = 1 \nleq 0.5

If n = 5 , then

P(Y) = \frac{365!}{(365)^{5} (365-5)! } = \frac{365!}{(365)^{5} (360)! }  = 0.97 \nleq 0.5

If n = 10 , then

P(Y) = \frac{365!}{(365)^{10} (365-10)! } = \frac{365!}{(365)^{10} (354)! }  = 0.88 \nleq 0.5

If n = 15 , then

P(Y) = \frac{365!}{(365)^{15} (365-15)! } = \frac{365!}{(365)^{15} (350)! }  = 0.75 \nleq 0.5

If n = 20 , then

P(Y) = \frac{365!}{(365)^{20} (365-20)! } = \frac{365!}{(365)^{20} (345)! }  = 0.588 \nleq 0.5

If n = 22 , then

P(Y) = \frac{365!}{(365)^{22} (365-22)! } = \frac{365!}{(365)^{22} (343)! }  = 0.52 \nleq 0.5

If n = 23 , then

P(Y) = \frac{365!}{(365)^{23} (365-23)! } = \frac{365!}{(365)^{23} (342)! }  = 0.49 \nleq 0.5

∴ we get

Number of students should be in class in order to have this probability above 0.5 = 23

5 0
3 years ago
What number must you add to complete the square? x^2-7
lys-0071 [83]

Answer:

12.25

Step-by-step explanation:

To find the c value in the equation, we need to divide b by two and then square it.

b= -7 and -7/2 is 3.5

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3 0
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Complete the pattern 3,665 3,765 3,865 _____. _____. _____.
rosijanka [135]
As we can see from the above the pattern is:

-In ascending order.

-A value of 100 is added to each new number.

The rest of the pattern will be:

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Therefore,the answer is: 3965. 4065. 4165.

Hope it helps!
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A group of students were doing a science experiment in which they measured wind speed and temperature. The first day the wind sp
Mama L [17]
So the difference of speed will be 25-17 = 8 mph  the last one choice 

hope helped 
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4 years ago
Abed says he has written a system of two linear equations that has an infinite number of solutions. One of the equations of the
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Answer:

to have an infinite number of solutions, the lines have to be on top of each other.....basically, they have to be the same line.

y = 3x - 1

-3x + y = -1.....multiply by -1

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Step-by-step explanation:

6 0
3 years ago
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