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3 years ago

One-half (1/2) hour equals______________.

Mathematics

2 answers:

ozzi3 years ago

5 0

1 hour = 60 minutes = 3600 seconds = 1/24 day

Divide every number above by 2 to get 1/2 hour and its equivalent numbers.

1/2 hour = 30 minutes = 1800 seconds = 1/12 day

adell [148]3 years ago

3 0

One-half hour is equal to 30 min.

One hour is equal to 60 min and half of that is 30. Hope this helps

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Step-by-step explanation:

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3 years ago

What is -7.08 as a fraction or mixed number

Basile [38]

-7.08 as a fraction would be -177 over 25

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3 years ago

Pedro has a bag of flour that weighs 9/10 pound. he uses 2/3 of the bag to make gravy. how many pounds of flour does pedro use t

BARSIC [14]

Here is the solution of the given problem above.
First, let's analyze the question.
Given: 1 bag = 9/10 pound
2/3 bag = ? pound
What we are going to do is to divide 9/10 pound to 3.
so 9/10 divided by 3 and we get 9/30 and to simplify that, 3/10.
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7 0

3 years ago

Read 2 more answers

A coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces. A random sample of 15 co

Luda [366]

Answer:

We conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

Step-by-step explanation:

We are given that a coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces.

A random sample of 15 containers were weighed and the mean weight was 31.8 ounces with a sample standard deviation of 0.48 ounces.

Let $\mu$ = mean weight of coffee in its containers.

SO, Null Hypothesis, $H_0$ : $\mu \geq$ 32 ounces {means that the mean weight of coffee in its containers is at least 32 ounces}

Alternate Hypothesis, $H_A$ : $\mu$ < 32 ounces {means that the mean weight of coffee in its containers is less than 32 ounces}

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

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

where, $\bar X$ = sample mean weight = 31.8 ounces

s = sample standard deviation = 0.48 ounces

n = sample of containers = 15

So, the test statistics = $\frac{31.8 -32}{\frac{0.48}{\sqrt{15} } }$ ~ $t_1_4$

= -1.614

The value of t test statistics is -1.614.

Now, at 0.01 significance level the t table gives critical value of -2.624 at 14 degree of freedom for left-tailed test.

Since our test statistic is more than the critical value of t as -1.614 > -2.624, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

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3 years ago