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3 years ago

The diagram to the right shows a reflection across a line. Find the value of each variable.

Mathematics

2 answers:

puteri [66]3 years ago

7 0

From the diagram, we can conclude that 5x = 15, 4y = 8, and 2z - 1 = 24.From that, we solve and get x = 3, y = 2, and z = 12.5

Hope this helps!

nydimaria [60]3 years ago

4 0

5x = 15

x = 3 <===== x is here

4y = 8

y = 2 <===== y is here

2z - 1 = 24

2z = 25

z = 12.5 <===== z is here

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garri49 [273]

Answer:

Step-by step

What do we already know about the equation?

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3 years ago

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3 years ago

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alexgriva [62]

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3 years ago

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valentina_108 [34]

Answer:

$\boxed { \frac{7}{5}y } \to \: \boxed {y+ \frac{2}{5}y }$

$\boxed { \frac{3}{5}y } \to \: \boxed {y - \frac{2}{5}y }$

Step-by-step explanation:

We need to simplify

$y + \frac{2}{5}y$

We collect LCM to get;

$\frac{5y + 2y}{5} = \frac{7y}{5}$

Therefore:

$\boxed { \frac{7}{5}y } \to \: \boxed {y+ \frac{2}{5}y }$

Also we need to simplify:

$y - \frac{2}{5}y$

We collect LCM to get;

$y - \frac{2}{5}y = \frac{5y - 2y}{5} = \frac{3}{5} y$

Therefore

$\boxed { \frac{3}{5}y } \to \: \boxed {y - \frac{2}{5}y }$

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3 years ago

In 1982 Abby’s mother scored at the 93rd percentile in the math SAT exam. In 1982 the mean score was 503 and the variance of the

oksian1 [2.3K]

Answer:

The percentle for Abby's score was the 89.62nd percentile.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation(which is the square root of the variance) $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Abby's mom score:

93rd percentile in the math SAT exam. In 1982 the mean score was 503 and the variance of the scores was 9604.

93rd percentile. X when Z has a pvalue of 0.93. So X when Z = 1.476.

$\mu = 503, \sigma = \sqrt{9604} = 98$