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3 years ago

Helpp PLZZZ due tomorrow

Mathematics

2 answers:

Dima020 [189]3 years ago

8 0

3. 12
4. 6
5. 16
6. 12
7. 14400
8. 10000
9. 180
10. 104
11. 6 pints

Ira Lisetskai [31]3 years ago

6 0

Its 12 6 16 12 pretty sure

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Write and solve a proportion that the teacher can use to estimate how many students in the whole school would choose the aquariu

Anon25 [30]

Answer:

Never gonna

Step-by-step explanation:

give you up,

Let you down,

make you cry

Run around and desert you

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2 years ago

What is the solution to the system of equations represented by the two equations?

nexus9112 [7]

Just substitute the first equation into the second and solve for x: 32x=−12x+4⟹2x=4⟹x=2Now, since y = 3/2 * x, we have y = 3 and we are done.

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3 years ago

Breyers is a major producer of ice cream and would like to test the hypothesis that the average American consumes more than 17 o

zhenek [66]

Answer:

The p-value would be 0.0939

Step-by-step explanation:

We can set a standard t-test for the Null Hypothesis that $H_0: 18.2\geq 17$

The test statistic then takes the form

$t=\frac{18.2-17}{3.9/\sqrt{15}}=1.317$

with this value we then can calculate the probability that is left to the right of this value $P(X>1.317)$ . From theory we know that t follows a standard normal distribution. Then $P(X>1.317)=0.0939$ which is smaller than the p-value set by Breyers of 0.10

3 0

3 years ago

Assume that the probability of a defective computer component is 0.02. Components arerandomly selected. Find the probability tha

solmaris [256]

Answer:

0.0177 = 1.77% probability that the first defect is caused by the seventh component tested.

The expected number of components tested before a defective component is found is 50, with a variance of 0.0208.

Step-by-step explanation:

Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested.

First six not defective, each with 0.98 probability.

7th defective, with 0.02 probability. So

$p = (0.98)^6*0.02 = 0.0177$

0.0177 = 1.77% probability that the first defect is caused by the seventh component tested.

Find the expected number and variance of the number of components tested before a defective component is found.

Inverse binomial distribution, with $p = 0.02$

Expected number before 1 defective(n = 1). So

$E = \frac{n}{p} = \frac{1}{0.02} = 50$

Variance is:

$V = \frac{np}{(1-p)^2} = \frac{0.02}{(1-0.02)^2} = 0.0208$

The expected number of components tested before a defective component is found is 50, with a variance of 0.0208.

5 0

3 years ago

A boy purchased (bought) a party-length sandwich 51 in. long. He wants to cut it into three pieces so that the middle piece is

shepuryov [24]

Answer:

Step-by-step explanation:

we have a 51 in sandwich

let a piece be x

we have x= shorter piece

another piece is x+6

another piece (x+6)-9=x-3

x+x+6+x-3=51

3x=51-3

3x=48

x=48/3=16

x=16, longer piece is 16+6=22, shorter piece =10

6 0

2 years ago