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kaheart [24]
3 years ago
12

PLZ HELP ASAPThe scale factor of ____ was applied to the first triangle

Mathematics
1 answer:
pashok25 [27]3 years ago
5 0

The scale factor of 4 was applied to the first triangle. 6 was divided by 4 to get 1.5

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Answer:

Length = Width * 3

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Perimeter = (L * 2) + (W * 2)

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Y=5
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15. = slope would be 7 and y-int would be -4
16. = slope would be -2/5 and y-int would be 0
17. = doesn’t have a y variable

slope intercept form is y=mx+b with the m being the slope and the b being the y-int. in some cases where the equation is not in this form you have to change it so it is in that form by using opposite operations
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3 years ago
Can someone help me please?? in Algblra2 - Variations, Progression, and Theorems
VashaNatasha [74]
If (y-1) is a factor of f(y), f(y)=0 when y=1.  So if you find that f(1)=0, then (y-1) is a factor of f(y).

f(y)=y^3-9y^2+10y+5

f(1)=1-9+10+5=7

Since f(1)=7, (y-1) is not a factor.
5 0
3 years ago
g Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distrib
algol13

Answer:

The probability that the sample mean would be at least $39000 is of 0.8665 = 86.65%.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean $39725 and standard deviation $7320.

This means that \mu = 39725, \sigma = 7320

Sample of 125

This means that n = 125, s = \frac{7320}{\sqrt{125}}

The probability that the sample mean would be at least $39000 is about?

This is 1 subtracted by the pvalue of Z when X = 39000. So

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

By the Central Limit Theorem

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

Z = \frac{39000 - 39725}{\frac{7320}{\sqrt{125}}}

Z = -1.11

Z = -1.11 has a pvalue of 0.1335

1 - 0.1335 = 0.8665

The probability that the sample mean would be at least $39000 is of 0.8665 = 86.65%.

4 0
2 years ago
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