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topjm [15]
3 years ago
10

ΔXYZ is translated 4 units up and 3 units left to yield ΔX′Y′Z′. What is the distance between any two corresponding points on ΔX

YZ and ΔX′Y′Z′ ?
a. 25 units
b. 7 units
c. 5 units
d. `sqrt(7)` units
Mathematics
1 answer:
kherson [118]3 years ago
3 0

c. 5 units_____________________________________________________________________________________________________________________

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Answer:

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Someone answer the whole thing fast plz
Brilliant_brown [7]

Answer:

y = 2x + 54

Step-by-step explanation:

slope of equation:

(y2-y1)/(x2-x1)

= (72-58)/(9-2)

= 14/7

= 2

y = 2x + b

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6 0
3 years ago
Suppose the weights of the Boxers at this club are Normally distributed with a mean of 166 pounds and a standard deviation of 5.
lesya [120]

Answer:

0.1994 is the required probability.      

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 166 pounds

Standard Deviation, σ = 5.3 pounds

Sample size, n = 20

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

Standard error due to sampling =

=\dfrac{\sigma}{\sqrt{n}} = \dfrac{5.3}{\sqrt{20}} = 1.1851

P(sample of 20 boxers is more than 167 pounds)

P( x > 167) = P( z > \displaystyle\frac{167 - 166}{1.1851}) = P(z > 0.8438)

= 1 - P(z \leq 0.8438)

Calculation the value from standard normal z table, we have,  

P(x > 167) = 1 - 0.8006= 0.1994 = 19.94\%

0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds

3 0
2 years ago
Simplify:
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15x4y2 letter B is your answer
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2 years ago
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