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lina2011 [118]
2 years ago
5

In an enlargement PQ P1Q1|PQ|=3cm and|P1Q1|=15cm.Calculate the scale of the enlargement. ​

Mathematics
1 answer:
natulia [17]2 years ago
7 0

Answer:

374795-57770

Step-by-step explanation:

Abby uses

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\bf \sqrt{\sqrt{x-5}+x}=5\leftarrow \textit{squaring both sides}
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\sqrt{x-5}+x=25\implies \sqrt{x-5}=25-x\leftarrow \textit{squaring both sides}
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x-5=(25-x)^2\implies x-5=625-50x+x^2
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0=x^2-51x+630\implies 0=(x-30)(x-21)

and surely you'd know what the roots are

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3 years ago
What is the measure of Q?
SVETLANKA909090 [29]

Answer:

<Q = 30

Step-by-step explanation:

2x + 70 + x - 10 = 180

3x + 60 = 180

3x = 120

x = 40

x - 10

40 - 10

30

<Q = 30 because of vertical angles

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3 years ago
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Just makes a punch that is 1/4 cranberry juice which two fractions are equivalent to 1/4
viva [34]
2/8 and 4/16 are equivalent to 1/4

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4 years ago
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How to solve exponential equations
Anettt [7]
Is there a specific question?

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4 years ago
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Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distribut
defon

Answer:

0.8665 = 86.65% probability that the sample mean would be at least $39000

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}} = 654.72

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}{654.72}

Z = -1.11

Z = -1.11 has a pvalue of 0.1335

1 - 0.1335 = 0.8665

0.8665 = 86.65% probability that the sample mean would be at least $39000

4 0
3 years ago
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