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devlian [24]
3 years ago
6

Write each number in standard form 9.03 x 10² 7.89 x 10³ 4.115 x 10⁵

Mathematics
1 answer:
mars1129 [50]3 years ago
3 0
9.03 x 100
7.89x1000
4.115x100000
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svetlana [45]

Answer:

$4,425.7

Step-by-step explanation:

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a wheelbarrow can hold 2.4 pounds of cement at a time. if it takes samantha 0.5 trips to transport all of his cement, how much c
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Answer:

1.2 pounds

Step-by-step explanation:

if it only takes her half a trip then she would have half the amount of cement in her wheelbarrow

4 0
3 years ago
Steve ran 1 lap in 7 minutes. At this rate, how far would he run in 42 minutes?
VARVARA [1.3K]

Answer:

6 laps

Step-by-step explanation:

46 Divided by 7 is 6

3 0
2 years ago
Read 2 more answers
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.

8 0
3 years ago
Please help, thank you :)
Delicious77 [7]

X + 4 - 5/X

Step-by-step explanation:

I believe that's what it is but I could be wrong it could also be 1/x

6 0
3 years ago
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