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3 years ago

Find m<6 if m<8 = 112°

Mathematics

1 answer:

grin007 [14]3 years ago

5 0

<h3>Answer: 112 degrees</h3>

Reason:

Angle 6 and angle 8 are corresponding angles because they are in the same southeast corner of their four-corner angle configuration. Corresponding angles are congruent when we have parallel lines like this.

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What is the slope of the line y = 3x + 5?<br> -3<br> -5<br> 5<br> 3

melomori [17]

Answer:

The slope is 3

Step-by-step explanation:

The number before the x is always the slope

5 0

3 years ago

Without doing any computation, decide which has a higher probability, assuming each sample is from a population that is normally

Nana76 [90]

Answer: The probability in (b) has higher probability than the probability in (a).

Explanation:

Since we're computing for the probability of the sample mean, we consider the z-score and the standard deviation of the sampling distribution. Recall that the standard deviation of the sampling distribution approximately the quotient of the population standard deviation and the square root of the sample size.

So, if the sample size higher, the standard deviation of the sampling distribution is lower. Since the sample size in (b) is higher, the standard deviation of the sampling distribution in (b) is lower.

Moreover, since the mean of the sampling distribution is the same as the population mean, the lower the standard deviation, the wider the range of z-scores. Because the standard deviation in (b) is lower, it has a wider range of z-scores.

Note that in a normal distribution, if the probability has wider range of z-scores, it has a higher probability. Therefore, the probability in (b) has higher probability than the probability in (a) because it has wider range of z-scores than the probability in (a).

5 0

3 years ago

Assume Y=1+X+u, where X, Y, and u=v+X are random variables, v is independent of X; E(v)=0, Var(v)=1, E(X)=1, and

kobusy [5.1K]

Answer:

a) $E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1+$

b) $E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2$

c) $E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2$

d) $E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6$

e) $E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1$

f) $E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3$

g) $E(u) = E(v) +E(X) = 0+1=1$

h) E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]

Step-by-step explanation:

For this case we know this:

$Y = 1+X +u$

$u = v+X$

with both Y and u random variables, we also know that:

$[tex] E(v) = 0, Var(v) =1, E(X) = 1, Var(X)=2$

And we want to calculate this:

Part a

$E(u|X=1)= E(v+X|X=1)$

Using properties for the conditional expected value we have this:

$E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1$

Because we assume that v and X are independent

Part b

$E(Y| X=1) = E(1+X+u|X=1)$

If we distribute the expected value we got:

$E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2$

Part c

$E(u|X=2)= E(v+X|X=2)$

Using properties for the conditional expected value we have this:

$E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2$

Because we assume that v and X are independent

Part d

$E(Y| X=2) = E(1+X+u|X=2)$

If we distribute the expected value we got:

$E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6$

Part e

$E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1$

Part f

$E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3$

Part g

$E(u) = E(v) +E(X) = 0+1=1$

Part h

E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]

8 0

3 years ago

Chen is bringing fruit and veggies to serve at an afternoon meeting. He spends a total of $28.70 on 5 pints of cut veggies and 7

ololo11 [35]

Given:
total cost : 28.70
cost per pint of fruit : 2.85

Formula:
5v + 7f = 28.70
5v + 7(2.85) = 28.70
5v + 19.95 = 28.70
5v = 28.70 - 19.95
5v = 8.75
5v/5 = 8.75/5

v = 1.75 cost per pint of veggies.

to check:
v = 1.75 ; f = 2.85 ; total cost = 28.70

5v + 7f = 28.70
5(1.75) + 7(2.85) = 28.70
8.75 + 19.95 = 28.70
28.70 = 28.70

7 0

3 years ago

Janie was given (-2, 3) as the endpoint of a line segment and (0,1) as its midpoint, which ordered pair should Janie choose as h

MaRussiya [10]

When given two end points of a segment say for example (X1,Y1) and (X2,Y2) to get the midpoint of the line or the segment we use the formula,
midpoint =( (X1+X2)/2 , (Y1+Y2)/2 )
therefore in our case the midpoint is (0,1)
hence, (-2 + X2)/ 2 = 0, thus X2 = 2
and (3 + Y2)/ 2= 1 , thus Y2 = -1
Therefore the other end point will be (2,-1). Thus none of the above paper can be used as the other end point.

8 0

3 years ago