HELPPP FIND THE AREA !!

Mathematics

1 answer:

Rashid [163]2 years ago

7 0

Answer:

To find the area, you have to multiply its height by its width.

Step-by-step explanation:

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Quadrilateral ABCD has vertices A(-3, 4), B(1, 3), C(3, 6), and D(1, 6). Match each set of vertices of quadrilateral EFGH with t

Sati [7]

Each transformation will give these following coordinates

Translation 7 units right gives E(4, 4), F(8, 3), G(10, 6), and H(8, 6) as shown in brown in the diagram below

Reflection on the y-axis gives E(3, 4), F(-1, 3), G(-3, 6) and H(-1, 6) as shown in green in the diagram below

Reflection on the x-axis gives E(-3, -4), F(1, -3), G(3, -6) and H(1, -6) as shown in red in the diagram below

Translation 5 units down gives E(-3, -1), F(1, -2, G(3, 1) and H(1, 1) as shown in purple in the diagram below

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3 years ago

An elementary school class ran 1 mile in an average of 11 minutes with a standard deviation of 3 minutes. Rachel, a student in t

Natali5045456 [20]

Answer:

a) Kenji's time had a lower z-score, which means that he is better compared to other runners on his level, and better to his peers compared to Nedda.

b) Rachel, because her time had the lowest z-score.

Step-by-step explanation:

Z-score:

In a set with mean $\mu$ and standard deviation $\sigma$ , 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.

In this question:

The lower the time, the better the runner is. So whoever's time has a lower z-score is a better runner. So initially, we find the z-score of each student's time.

An elementary school class ran 1 mile in an average of 11 minutes with a standard deviation of 3 minutes. Rachel, a student in the class, ran 1 mile in 8 minutes.

This means that for Rachel's time, we have that $X = 8, \mu = 11, \sigma = 3$ . So

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

$Z = \frac{8 - 11}{3}$

$Z = -1$

A junior high school class ran 1 mile in an average of 9 minutes, with a standard deviation of 2 minutes. Kenji, a student in the class, ran 1 mile in 8.5 minutes.

This means that for Kenji's time, we have that $X = 8.5, \mu = 9, \sigma = 2$ . So

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

$Z = \frac{8.5 - 9}{2}$

$Z = -0.25$

A high school class ran 1 mile in an average of 7 minutes with a standard deviation of 4 minutes. Nedda, a student in the class, ran 1 mile in 8 minutes.

This means that for Nedda's time, we have that $X = 8, \mu = 7, \sigma = 4$ . So