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Vilka [71]
3 years ago
8

What can you tell about the mean of each distribution? (Please hurry!)

Mathematics
1 answer:
Oliga [24]3 years ago
7 0

Answer:its c

Step-by-step explanation:

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Please help me <br> 30 points <br> Show you work
GenaCL600 [577]

Answer:

(a) A translation is a rigid transformation. How does this statement support line s being parallel to line r?

Because ALL of the points of line r move the same distance in the same direction, then the inclination of the line s doesn't change with respect line r.

(b) Write and expression for the slope of line r:

(k-n)/(j-m)

(c) Write and expression for the slope of line s:

(k-n)/(j-m)

(d) Line q is a vertical is a vertical translation of line s 3 units down. P'' is the image of P'. What are the coordinates of P''?

The coordinates of P'' are:

P''=(m,n+ϴ-3)

Step-by-step explanation:

(a) A translation is a rigid transformation. How does this statement support line s being parallel to line r?

Because ALL of the points of line r move the same distance in the same direction, then the inclination of the line s doesn't change with respect line r.


(b) Write and expression for the slope of line r.

Line r goes through the points:

P=(m,n)=(xp,yp)→xp=m, yp=n

Q=(j,k)=(xq,yq)→xq=j, yq=k

Slope of line r is:

mr=(yq-yp)/(xq-xp)

Replacing the coordinates, the expression for the slope of line r is:

mr=(k-n)/(j-m)


(c) Write and expression for the slope of line s.

Line s goes through the points:

P'=(m,n+ϴ)=(xp',yp')→xp'=m, yp'=n+ϴ

Q'=(j,k+ϴ)=(xq',yq')→xq'=j, yq'=k+ϴ

Slope of line s is:

ms=(yq'-yp')/(xq'-xp')

Replacing the coordinates, the expression for the slope of line s is:

ms=[(k+ϴ)-(n+ϴ)]/(j-m)

Eliminating the parentheses in the numerator:

ms=[k+ϴ-n-ϴ]/(j-m)

Simplifyng:

ms=(k-n)/(j-m)


(d) Line q is a vertical is a vertical translation of line s 3 units down. P'' is the image of P'. What are the coordinates of P''?

3 units down, then change only the ordinate y:

P''=(xp',yp'-3)

Replacing xp' and yp':

P''=(m,n+ϴ-3)


7 0
3 years ago
Read 2 more answers
Can somebody help me?
Degger [83]
A percentage
Have a good dayyyyyy
5 0
3 years ago
Answer the question for a normal random variable x with mean μ and standard deviation σ specified below. (Round your answer to o
vladimir2022 [97]

Answer:

x = 63.6

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X and also the area to its left. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X, which is the area to its right.

In this problem, we have that:

\mu = 38, \sigma = 11

Find a value of x that has area 0.01 to its right

This is x when Z has a pvalue of 1-0.01 = 0.99. So it is X when Z = 2.3267.

Z = \frac{X - \mu}{\sigma}

2.3267 = \frac{X - 38}{11}

X - 38 = 11*2.3267

X = 63.6

So x = 63.6

6 0
3 years ago
A certain plant runs three shifts per day. Of all the items produced by the plant, 50% of them are produced on the first shift,
Snezhnost [94]

Answer:

0.2941 = 29.41% probability that it was manufactured during the first shift.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Defective

Event B: Manufactured during the first shift.

Probability of a defective item:

1% of 50%(first shift)

2% of 30%(second shift)

3% of 20%(third shift).

So

P(A) = 0.01*0.5 + 0.02*0.3 + 0.03*0.2 = 0.017

Probability of a defective item being produced on the first shift:

1% of 50%. So

P(A \cap B) = 0.01*0.5 = 0.005

What is the probability that it was manufactured during the first shift?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.005}{0.017} = 0.2941

0.2941 = 29.41% probability that it was manufactured during the first shift.

6 0
3 years ago
Integration of the following function
DerKrebs [107]

Answer: None of the above

Step-by-step explanation:

Given

\frac{\mathrm{d} \bar{c}}{\mathrm{d} x}=-6x^{-2}+\frac{1}{6}

d\bar{c}=\left (  -6x^{-2}+\frac{1}{6}\right )dx

Now, integrating both sides

\int d\bar{c}=\int \left (  -6x^{-2}+\frac{1}{6}\right )dx

\bar{c}=-6\frac{x^{-2+1}}{-2+1}+\frac{1}{6}x+k_1

\bar{c}=6x^{-1}+\frac{1}{6}x+k_1

6 0
3 years ago
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