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3 years ago

Please solve this i have no idea how to do this

Mathematics

2 answers:

nika2105 [10]3 years ago

5 0

If its at what speed is she jogging then this equals the slope of the line

which is 4 miles / 40 minuits = 4 / 2/3 mph = 4*3 / 2 = 6 mph

Arisa [49]3 years ago

4 0

Hihi!

I would say that for every 5 minutes she has jogged half a mile! So just keep that in mind! If you need anymore help please be sure to ask!

I hope I helped!
-Jailbaitasmr

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Evaluate: 10 x [(6 + 4) = 2]

svetlana [45]

Answer:

You have to pay close attention to the Order of Operations.

10(6+4)/2. Grouping symbols first.

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Step-by-step explanation:

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3 years ago

The glacier moves 5 centimeters in 8 days. How many days does it take the glacier to move 1 centimeter?

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4 years ago

A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past

Vadim26 [7]

Answer:

We conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

Step-by-step explanation:

We are given that a particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8000.

From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9800 miles.

Let $\mu$ = average miles for deluxe tires

So, Null Hypothesis, $H_0$ : $\mu \geq$ 50,000 miles {means that deluxe tire averages at least 50,000 miles before it needs to be replaced}

Alternate Hypothesis, $H_A$ : $\mu$ < 50,000 miles {means that deluxe tire averages less than 50,000 miles before it needs to be replaced}

The test statistics that will be used here is One-sample z test statistics as we know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean lifespan = 46,500 miles

$\sigma$ = population standard deviation = 8000 miles

n = sample of tires = 28

So, test statistics = $\frac{46,500-50,000}{\frac{8000}{\sqrt{28} } }$

= -2.315

The value of the test statistics is -2.315.

Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z as -2.315 < -1.6449, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

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