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Vikki [24]
2 years ago
6

Find the slope of the line that passes through (8,9) and (1, 3).

Mathematics
2 answers:
vlabodo [156]2 years ago
4 0
Rise (y) over run (x)
9-3/8-1
6/7
slope is 6/7
Finger [1]2 years ago
3 0

Answer:

m=6/7

Step-by-step explanation:

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Calvin had 30 minutes in time-out. For the first 23 1/3 minutes, Calvin counted spots on the ceiling. For the rest of the time,
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Answer: \frac{20}{3}\ minutes or 6 \frac{2}{3}\ minutes

Step-by-step explanation:

For this exercise you can convert the mixed number to an improper fraction:

1. Multiply the whole number part by the denominator of the fraction.

2. Add the product obtained and the numerator of the fraction (This will be the new numerator).

3. The denominator does not change.

Then:

23\frac{1}{3}= \frac{(23*3)+1}{3}= \frac{70}{3}\ minutes

You know that he had 30 minutes in time-out, he counted spots on the ceiling for \frac{70}{3} minutes and the rest of the time he made faces at his stuffed tiger.

Then, in order to calculate the time Calvin spent making faces at his stuffed tiger, you need to subract 30 minutes and \frac{70}{3} minutes:

30\ min-\frac{70}{3}=(\frac{3(30)-70}{3})=\frac{20}{3}\ minutes or 6 \frac{2}{3}\ minutes

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4.44 infinet

Step-by-step explanation:

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A child walks 5.0 meters north, then 4.0 meters east, and finally 2.0 meters south. What is the magnitude of the resultant displ
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Answer:

Answer = √(0.301 × 0.699 / 83) ≈ 0.050

A 68 percent confidence interval for the proportion of persons who work less than 40 hours per week is (0.251, 0.351), or equivalently (25.1%, 35.1%)

Step-by-step explanation:

√(0.301 × 0.699 / 83) ≈ 0.050

We have a large sample size of n = 83 respondents. Let p be the true proportion of persons who work less than 40 hours per week. A point estimate of p is  because about 30.1 percent of the sample work less than 40 hours per week. We can estimate the standard deviation of  as . A  confidence interval is given by , then, a 68% confidence interval is , i.e., , i.e., (0.251, 0.351).  is the value that satisfies that there is an area of 0.16 above this and under the standard normal curve.The standard error for a proportion is √(pq/n), where q=1−p.

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