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3 years ago

List 5 fractions equivalent to 1/2

2 answers:

8 0

$\frac{2}{4} = \frac{1}{2}$
$\frac{3}{6} = \frac{1}{2}$
$\frac{4}{8} = \frac{1}{2}$
$\frac{6}{12} = \frac{1}{2}$
$\frac{12}{24} = \frac{1}{2}$

jolli1 [7]3 years ago

7 0

2/4, 3/6, 4/8, 5/10, 6/12, etc.

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Whats equivalent to 3.5 x 100.

ioda

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350

Step-by-step explanation:

since there are 2 zeros in "100", move the decimal over 2 places

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A rectangle has a height of 777 and a width of 2x^2-32x

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<h2>Explanation:</h2><h2></h2>

Hello! Remember you have to write clear questions in order to get good and exact answers. Here I'll assume the data given in a comment above. So:

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3 years ago

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3 years ago

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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 $\mu$ = <u><em>true average estimated calorie content in the population sampled.</em></u>

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 <u>One-sample t test statistics</u> 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, <u><em>the test statistics</em></u> = $\frac{193-153}{\frac{88}{\sqrt{58} } }$ ~ $t_5_7$

= 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%. <u>Now, at 0.05 significance level the t table gives critical value of 1.6725 at 57 degree of freedom for right-tailed test.</u>

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 <u>we reject our null hypothesis</u>.

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