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Kryger [21]
2 years ago
9

Find the scale factor from DEF to ABC

Mathematics
1 answer:
Serjik [45]2 years ago
8 0

Answer:

³/2 (1.5)

Step-by-step explanation:

∆DEF was enlarged to get ∆ABC. The scale factor would be:

The ratio of any of the side length of ∆ABC to the corresponding side length of ∆DEF.

Thus,

BC:EF = 12:8

= 12/8

= 3/2 = 1.5

Scale factor from ∆DEF to ∆ABC is ³/2 or 1.5

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Answer:

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Step-by-step explanation:

(-4,3),(1,-6),(-8,-1),(1,9)

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*hope that makes sense*

4 0
2 years ago
What is the next term of 54, 36, 24
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Answer:

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Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave
Colt1911 [192]
Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \  \frac{a-\mu}{\sigma}\right)

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\  \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\  \\ =1-P(z\ \textless \ -0.5410) \\  \\ =1-0.29426=\bold{0.7057}



Part B:

</span>The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

</span>P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}
7 0
3 years ago
Please help :) for this question
madreJ [45]

The awnser is D to your question

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