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lesya692 [45]
2 years ago
15

P + 3.89 = 19.74 Have a good weekend

Mathematics
1 answer:
m_a_m_a [10]2 years ago
7 0

Answer:

15.85

Step-by-step explanation:

P + 3.89 = 19.74

Subtract 3.89 from both sides to isolate the variable.

P = 15.85

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What is the solution to this system of equations?
labwork [276]

Answer:

C

Step-by-step explanation:

C. x=7, y=5

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1 year ago
5th Grade Math
LUCKY_DIMON [66]
516 is the dividend, 3 is the divisor, 172 would be the quotient.

3 goes into 5 once with a 2 remainder, that 2 turns into 21 because of the 1 in 516 which 3 goes into seven times with a remainder of zero turning into 6, which 3 goes into twice, making the quotient 172.

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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
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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 = <u><em>mean weight of coffee in its containers.</em></u>

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 <u>One-sample t-test statistics</u> 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, <u><em>the test statistics</em></u>  =  \frac{31.8 -32}{\frac{0.48}{\sqrt{15} } }  ~ t_1_4  

                                      =  -1.614

The value of t test statistics is -1.614.

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

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 <u><em>we fail to reject our null hypothesis</em></u>.

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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Which angles are supplementary to 14 ? select all that apply.
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