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3 years ago

Determine the measures of angle ACB, angle DCE, & Angle BCD.

Mathematics

1 answer:

ruslelena [56]3 years ago

7 0

Answer: ∠ACB= 60°, ∠DCE= 65°, ∠BCD= 55°

Step-by-step explanation:

1) Since ΔABC is an equilateral, we know ∠ACB= 60°

2) Since we know ΔCDE is an isosceles, we know that ∠DCE= 1/2 (180-50), which is 65°

3) We can add the 2 angles and then subtract them from 180 degrees to get ∠BCD= 55°

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Mr. Adams drove his delivery truck 151.2 miles during 2.4 days. He drove the same number of miles each day. How many miles did h

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Answer 63 miles
Explains 151.2 / 2.4 = 63

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3 years ago

In the Pythagorean equation, a and b represent the legs that create the right angle.

Alla [95]

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3 years ago

Read 2 more answers

4879.64 is 7% of what number?

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2 years ago

The amounts of electricity bills for all households in a city have a skewed probability distribution with a mean of $139 and a

ZanzabumX [31]

Answer:

P (within $6 of 4) = 0.9164

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 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.

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 $s = \frac{\sigma}{\sqrt{n}}$ .

Mean of $139 and a standard deviation of $30.

This means that $\mu = 139, \sigma = 30$

Random sample of 75 households

This means that $n = 75, s = \frac{30}{\sqrt{75}} = 3.464$

75 > 30, which means that the sampling distribution is approximately normal.

Find the probability that the mean amount of electric bills for a random sample of 75 households selected from this city will be within $6 of the population mean.

This is the pvalue of Z when X = 139 + 6 = 145 subtracted by the pvalue of Z when X = 139 - 6 = 133.

X = 145

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

By the Central Limit Theorem

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

$Z = \frac{145 - 139}{3.464}$

$Z = 1.73$

$Z = 1.73$ has a pvalue of 0.9582

X = 133

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

$Z = \frac{133 - 139}{3.464}$

$Z = -1.73$

$Z = -1.73$ has a pvalue of 0.0418

0.9582 - 0.0418 = 0.9164

P (within $6 of 4) = 0.9164

4 0

2 years ago

a rectangle is dilated using a scale factor of 6. The image us then dilated using a scale factor 1/3.What scale facotr could you

Mandarinka [93]

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3 years ago