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andrey2020 [161]
3 years ago
13

Ashton used 12.6 gallons of gasoline to drive his car on a weekend trip. He averaged 21.5 miles per gallon. About how many miles

did he travel?

Mathematics
1 answer:
Advocard [28]3 years ago
4 0

the answer is c




cdiwcdbbuibuiwvvbuuvvbueqvebqubepvvpqebuvrvbbruurivbpriuvbpeirubvpqieubviquebvhiebviubsidbv

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The scatter plot shows the time spent studying, x, and the midterm score, y, for each of 24 students.
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3 years ago
Suppose that the rate of return on stocks is normally dis-tributed with mean of 9% and a standard deviation of 3%. If I pick fiv
alisha [4.7K]

Answer:

The probability that at least two stocks will have a return of more than 12% is 0.1810.

Step-by-step explanation:

Let <em>X</em> = rate of return on stocks.

The random variable <em>X</em> follows a Normal distribution, N (9, 3²).

Compute the probability that a stock has rate of return more than 12% as follows:

P(X\geq 12)=1-P(X

**Use the <em>z</em> table for the probability.

The probability of a stock having rate of return more than 12% is 0.1587.

Now define a random variable <em>Y</em> as the number of stocks that has rate of return more than 12%.

The sample size of stocks selected is, <em>n</em> = 5.

The random variable <em>Y </em>follows a Binomial distribution.

The probability of a Binomial distribution is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0, 1, 2, ...

Compute the value of P (X ≥ 2) as follows:

P (X ≥ 2) = 1 - P (X < 2)

              = 1 - P (X = 0) - P (X = 1)

              =1-{5\choose 0}(0.1587)^{0}(1-0.1587)^{5-0}-{5\choose 1}(0.1587)^{1}(1-0.1587)^{5-1}\\=1-0.4215-0.3975\\=0.1810

Thus, the probability that at least two stocks will have a return of more than 12% is 0.1810.

6 0
3 years ago
Slope of (16, 8) (8, 4)
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Answer:

slope of (16,8) (8,4) is 1/2

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2 years ago
At Silver Gym, membership is $30 per month, and personal training sessions are $45 each. At Fit Factor, membership is $90 per mo
pshichka [43]
S(x)=30x+45
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3 years ago
Anyone know how to find this answer
Kaylis [27]
The answer is a i hope i helped
4 0
3 years ago
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