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1 year ago

ASAP PLS I NEED IT AND TY

Mathematics

2 answers:

mars1129 [50]1 year ago

8 0

<h2>Dilations and Scale Factors</h2>

When we're solving with scale factors, we either divide or multiply.

If the shape becomes larger, we multiply to get the new length.

If the shape becomes smaller, we divide to get the new length.

<h2>Solving the Question</h2>

We're given:

Original side length is 2.5 units
Scale factor = 6

Because the shape becomes larger, we have to multiply the original side length by the scale factor, 6, to get the new side length.

2.5 x 6 = 15

Therefore, the missing side length is 15 units.

<h2>Answer</h2>

15 units

jeka57 [31]1 year ago

7 0

Answer:

Step-by-step explanation:

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3 years ago

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arsen [322]

Answer:

A tree with a height of 6.2 ft is 3 standard deviations above the mean

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⇒ $1^s^t$ statement: A tree with a height of 5.4 ft is 1 standard deviation below the mean(FALSE)

an X value is found Z standard deviations from the mean mu if:

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

In this case we have: $\mu=5\ ft$ $\sigma=0.4\ ft$

We have four different values of X and we must calculate the Z-score for each

For X =5.4\ ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.4-5}{0.4}=1$

Therefore, A tree with a height of 5.4 ft is 1 standard deviation above the mean.

⇒ $2^n^d$ statement:A tree with a height of 4.6 ft is 1 standard deviation above the mean. (FALSE)

For X =4.6 ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{4.6-5}{0.4}=-1$

Therefore, a tree with a height of 4.6 ft is 1 standard deviation below the mean .

⇒ $3^r^d$ statement:A tree with a height of 5.8 ft is 2.5 standard deviations above the mean (FALSE)

For X =5.8 ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.8-5}{0.4}=2$

Therefore, a tree with a height of 5.8 ft is 2 standard deviation above the mean.

⇒ $4^t^h$ statement:A tree with a height of 6.2 ft is 3 standard deviations above the mean. (TRUE)

For X =6.2\ ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{6.2-5}{0.4}=3$

Therefore, a tree with a height of 6.2 ft is 3 standard deviations above the mean.

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3 years ago