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Zanzabum
3 years ago
8

Jamal ran 8 kilometers, sprinted 500 meters, and climbed bleachers for a distance of 1.5 kilometers during track practice. How m

any meters, in total, did he travel during practice?
Mathematics
2 answers:
Sindrei [870]3 years ago
7 0

Answer: He  travelled 10,000 meters during practice

Step-by-step explanation:

Hi, to answer this question we have to sum all the distances:

ran: 8 km

Sprinted: 500 meters

Climbed: 1.5 kilometers

We have to express all the distances in meters.

Since 1000 meters = 1 km

8 km x 1000= 8000 meters

1.5 km x 1000= 1500

Adding all the distances:

8000 m+ 1500m +500m = 10,000 meters

He  travelled 10,000 meters during practice

xeze [42]3 years ago
6 0
1 kilometer=1000 meters so the 8 kilometers he ran would be 8000 meters then theres the 500 he sprinted. he climbed bleachers for 1.5 kilometers which would equal 1500 meters so 8000+500+1500=10,000. He traveled 10,000 kilometers.
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Read 2 more answers
Hypothesis Testing for Means with Small Samples
Scorpion4ik [409]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is

X: volume of root beer in a Windsor Bottling Company can.

A sample of n=24 cans was taken and their contents measured, resulting:

X[bar]= 11.4 oz

S= 0.62 oz

Assuming that the variable has a normal distribution X~N(μ;σ²), the parameter of interest is the average contents of the root beer cans of the Windsor Bottling Company (μ)

The claim is that the population mean content of the cans is different from 12 oz, symbolically: μ ≠ 12

The statistical hypothesis (Null and alternative) have to be complementary, exhaustive and mutually exclusive. The null hypothesis is the "no change" hypothesis and always carries the "=" sign.

If the claim is μ ≠ 12, its complement is μ = 12, the expression carrying the "=" sign will be the null hypothesis and its complement will be the alternative hypothesis:

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α: 0.05

To test the population mean of this normal population, you have to apply a one sample t-test, with statistic:

t= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } ~t_{n-1}

t_{H_0}= \frac{11.4-12}{\frac{0.62}{\sqrt{24} } } = -4.74

This test is two-tailed, using the critical value approach, you have to determine two rejection regions. Meaning, you'll reject the null hypothesis to small values of the statistic or to high values of the statistic.

t_{n-1;\alpha /2}= t_{23;0.025}= -2.069

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The decision rule is:

If t_{H_0} ≤ -2.069 or if t_{H_0} ≥ 2.069, then you reject the null hypothesis.

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The value is less than the left critical value, the decision is to reject the null hypothesis.

Then you can say that with a 5% significance level, there is significant evidence to reject the null hypothesis, then the average amount of root beer of the Windsor Bottling Company is different from 12 oz, this means that the claim about the amount of root beer in the cans is correct.

I hope it helps!

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