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stepladder [879]
3 years ago
14

In quadrilateral TUVW, m<U = 94º, m <V = 51°, and m<w =67º, what is m<T°?

Mathematics
1 answer:
Vlada [557]3 years ago
4 0
The answer would be c) 129
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If A = 3r^2h^2, and (rh) increases by 100%, then the new A is how many times greater than the old A? PLEASE HELP ME!!! (Specify
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Answer: the answer is 3

Step-by-step explanation:

An increase of 100% in the value of (rh) means the value doubles. When that doubled value is squared, the new area is 4 times the old area.The question asks how many times GREATER the new A is than the old A. 4 times AS LARGE AS is a 300% INCREASE, which is 3 TIMES LARGER THAN.So the grammatically correct answer is 3 

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For a data set of the pulse rates for a sample of adult? females, the lowest pulse rate is 39 beats per? minute, the mean of the
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Answer:

a) The difference is of 35 beats per minute.

b) So 3.07 standard deviations below the mean.

c) Z = -3.07.

Step-by-step explanation:

Z-score:

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.

In this question:

Mean \mu = 74, standard deviation \sigma = 11.4

a. What is the difference between the pulse rate of 39 beats per minute and the mean pulse rate of the? females?

39 - 74 = -35

The difference is of 35 beats per minute.

b. How many standard deviations is that? [the difference found in part? (a)?

|Z| when X = 39. So

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

Z = \frac{39 - 74}{11.4}

Z = -3.07

|Z| = 3.07

So 3.07 standard deviations below the mean.

c. Convert the pulse rate of 39 beats per minutes to a z score.

From item b. above, Z = -3.07.

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