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2 years ago

8. Aaron was riding his bicycle to school at a rate of 18 miles an hour down the street. As he got

Mathematics

1 answer:

slavikrds [6]2 years ago

8 0

Answer:

Step-by-step explanation:

Aaron was riding forward, so the position graph should always be sloping upward, with his position always increasing away from home. Lastly, he slowed gradually as he got closer to the school, until he stopped at the bike rack, so the slope of X does not fit. Graph Z best fits this situation.

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Imagine your friend is writing an argument to convince adults that they should allow teens to own and use smartphones. Read her

Blababa [14]

I think it is the 2nd one

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2 years ago

Please help me asap! Could you please explain how you got the answer

svetoff [14.1K]

Answer:

B) 23

Step-by-step explanation:

Table: Algebra | No algebra | Total

Physics | 4 Step 1: 8 12

No physics |Step 2: 15

Total | 19 30

We look for the number of people who are in algebra and no physics or physics and no algebra.

1. 12 - 4 = 8

2. 19 - 4 = 15

3. We find the total of those numbers: 15 + 8 = 23

7 0

2 years ago

must a function that is decreasing over a given interval always be negative over that same interval explain

kolbaska11 [484]

check the picture below.

notice, a function that is decreasing, namely have a negative slope, doesn't have to be negative.

8 0

2 years ago

Read 2 more answers

Pick the description of a geometric object that best matches the real-world object:

svet-max [94.6K]

E ) a three-dimensional box-shaped figure, with six identical square faces

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2 years ago

Read 2 more answers

The article "Expectation Analysis of the Probability of Failure for Water Supply Pipes"† proposed using the Poisson distribution

oksian1 [2.3K]

Answer:

a. P(X ≤ 5) = 0.999

b. P(X > λ+λ) = P(X > 2) = 0.080

Step-by-step explanation:

We model this randome variable with a Poisson distribution, with parameter λ=1.

We have to calculate, using this distribution, P(X ≤ 5).

The probability of k pipeline failures can be calculated with the following equation:

$P(k)=\lambda^{k} \cdot e^{-\lambda}/k!=1^{k} \cdot e^{-1}/k!=e^{-1}/k!$

Then, we can calculate P(X ≤ 5) as:

$P(X\leq5)=P(0)+P(1)+P(2)+P(4)+P(5)\\\\\\P(0)=1^{0} \cdot e^{-1}/0!=1*0.3679/1=0.368\\\\P(1)=1^{1} \cdot e^{-1}/1!=1*0.3679/1=0.368\\\\P(2)=1^{2} \cdot e^{-1}/2!=1*0.3679/2=0.184\\\\P(3)=1^{3} \cdot e^{-1}/3!=1*0.3679/6=0.061\\\\P(4)=1^{4} \cdot e^{-1}/4!=1*0.3679/24=0.015\\\\P(5)=1^{5} \cdot e^{-1}/5!=1*0.3679/120=0.003\\\\\\P(X\leq5)=0.368+0.368+0.184+0.061+0.015+0.003=0.999$

The standard deviation of the Poisson deistribution is equal to its parameter λ=1, so the probability that X exceeds its mean value by more than one standard deviation (X>1+1=2) can be calculated as:

$P(X>2)=1-(P(0)+P(1)+P(2))\\\\\\P(0)=1^{0} \cdot e^{-1}/0!=1*0.3679/1=0.368\\\\P(1)=1^{1} \cdot e^{-1}/1!=1*0.3679/1=0.368\\\\P(2)=1^{2} \cdot e^{-1}/2!=1*0.3679/2=0.184\\\\\\P(X>2)=1-(0.368+0.368+0.184)=1-0.920=0.080$

4 0

2 years ago