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Yuliya22 [10]
3 years ago
11

1. Which of the following represents the relationship shown in the table below?

Mathematics
1 answer:
zhannawk [14.2K]3 years ago
4 0

1.) C is the correct choice

4.) B is the correct choice

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y=2^{x}

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The box plots show the data distributions for the number of customers who used a coupon each hour during a two-day
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The correct answer is A: Interquartile range

Step-by-step explanation:

Just took the test and this was the correct answer

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When an automobile is stopped with a roving safety patrol,each tire is checked for tire wear, and each headlight is checkedto se
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a) Joint ptobability distribution

\begin{pmatrix}  &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}

b) P(X<= 1 and Y <= 1) = P(X<= 1) * P(Y<=1) = 0.56

c) P(X + Y = 0)=0.3

d) P(X + Y <= 1)=0.53

Step-by-step explanation:

We have to construct the joint probability table with the marginal probabilities of X and Y.

X can take values from 0 to 2, and Y can take values from 0 to 4.

We can calculate each point of the joint probability as:

P(x,y)=P_x(x)*P_y(y)

Then, the joint probabilities are:

X=0 Y=0 Px=0.5 Px=0.6 P(0,0)=0.3

X=0 Y=1 Px=0.5 Px=0.1 P(0,1)=0.05

X=0 Y=2 Px=0.5 Px=0.05 P(0,2)=0.025

X=0 Y=3 Px=0.5 Px=0.05 P(0,3)=0.025

X=0 Y=4 Px=0.5 Px=0.2 P(0,4)=0.1

X=1 Y=0 Px=0.3 Px=0.6 P(1,0)=0.18

X=1 Y=1 Px=0.3 Px=0.1 P(1,1)=0.03

X=1 Y=2 Px=0.3 Px=0.05 P(1,2)=0.015

X=1 Y=3 Px=0.3 Px=0.05 P(1,3)=0.015

X=1 Y=4 Px=0.3 Px=0.2 P(1,4)=0.06

X=2 Y=0 Px=0.2 Px=0.6 P(2,0)=0.12

X=2 Y=1 Px=0.2 Px=0.1 P(2,1)=0.02

X=2 Y=2 Px=0.2 Px=0.05 P(2,2)=0.01

X=2 Y=3 Px=0.2 Px=0.05 P(2,3)=0.01

X=2 Y=4 Px=0.2 Px=0.2 P(2,4)=0.04

We can write it in the form of a matrix:

\begin{pmatrix}  &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}

b) From the joint probability P(X<= 1 and Y <= 1) is equal to

P(X\leq 1 \& Y \leq 1)=P(0,0)+P(0,1)+P(1,0)+P(1,1)\\\\P(X\leq 1 \& Y \leq 1)=0.3+0.05+0.18+0.03=0.56

We can calculate P(X<= 1) * P(Y<=1)

P_x(X\leq 1)=P_x(0)+P_x(1)=0.5+0.3=0.8\\\\P_y(Y\leq1)=P_y(0)+P_y(1)=0.6+0.1=0.7\\\\P_x(X\leq1)*P_y(Y\leq1)=0.8*0.7=0.56

Both calculations give the same result.

c) Probability of no violations

P(X+Y=0)=P(0,0)=0.3

d) P(X + Y <= 1)

P(X+Y \leq 1)=P(0,0)+P(0,1)+P(1,0)\\\\P(X+Y \leq 1)=0.3+0.05+0.18=0.53

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A. The difference in the heights of the bars will not appear as great.

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29.98

Step-by-step explanation:

You subtract the 69 from the 40.02 and ypu get 29.98

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