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3 years ago

What is 2 + 10? pls help!!!!!!

Mathematics

2 answers:

Aleks [24]3 years ago

4 0

Answer:

The answer is 12

Step-by-step explanation:

there you go

Jlenok [28]3 years ago

3 0

12 because if you take 2 and add it to 10 you get 12

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Omar rented a truck for one day. There was a base fee of $14.99, and there was an additional charge of 80 cents for each kilomet

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2 years ago

If the lengths of the sides of a triangle measure 5/4 m 3/2 m and 1/8 m, determine its perimeter.

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23/8

Step-by-step explanation:

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1/8 = 1/8

Then we just add the numerators and keep the denominator the same, which gets us 23/8 meters. Since 23/8 can be simplified, that's the answer.

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3 years ago

Which of the following habits is likely to reduce math anxiety?

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2 years ago

Unlike most packaged food products, alcohol beverage container labels are not required to show calorie or nutrient content. An a

zhenek [66]

Answer:

We conclude that the true average estimated calorie content in the population sampled exceeds the actual content.

Step-by-step explanation:

We are given that an article reported on a pilot study in which each of 58 individuals in a sample was asked to estimate the calorie content of a 12 oz can of beer known to contain 153 calories.

The resulting sample mean estimated calorie level was 193 and the sample standard deviation was 88.

Let $\mu$ = true average estimated calorie content in the population sampled.

So, Null Hypothesis, $H_0$ : $\mu$ $\leq$ 153 calories {means that the true average estimated calorie content in the population sampled does not exceeds the actual content}

Alternate Hypothesis, $H_A$ : $\mu$ > 153 calories {means that the true average estimated calorie content in the population sampled exceeds the actual content}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean estimated calorie level = 193 calories

s = sample standard deviation = 88

n = sample of individuals = 58

So, the test statistics = $\frac{193-153}{\frac{88}{\sqrt{58} } }$ ~ $t_5_7$

= 3.462

The value of t test statistics is 3.462.

Since, in the question we are not given the level of significance so we assume it to be 5%. Now, at 0.05 significance level the t table gives critical value of 1.6725 at 57 degree of freedom for right-tailed test.

Since our test statistic is more than the critical value of t as 3.462 > 1.6725, 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 the true average estimated calorie content in the population sampled exceeds the actual content.

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3 years ago