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2 years ago

(-2+8i)(3-10i) Step by Step

1 answer:

5 0

(-2+8i)x(3-10i)

-2 x 3 - 2 x (-10i) +8i x 3- 8i x 10i
-6+20i+24i - 80i ^2
-6+20i+24i - 80 x (-1)
-6+20i+24i + 80
Answer: 74+44i

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2 years ago

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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2 years ago