how many servings are in a bag of trail mix that weighs 11 1/2 ounces and each serving of trail miix weighs 3/4 ounces

Mathematics

1 answer:

Mariulka [41]3 years ago

8 0

let's firstly convert the mixed fraction to improper fraction, and then divide.

$\bf \stackrel{mixed}{11\frac{1}{2}}\implies \cfrac{11\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{23}{2}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{23}{2}\div \cfrac{3}{4}\implies \cfrac{23}{2}\cdot \cfrac{4}{3}\implies \cfrac{23}{3}\cdot \cfrac{4}{2}\implies \cfrac{23}{3}\cdot 2\implies \cfrac{46}{3}\implies 15\frac{1}{3}$

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andre [41]

Answer:

Option D) Yes, because the test statistic is -2.01

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 30 pound

Sample mean, $\bar{x}$ = 29.1 pounds

Sample size, n = 20

Alpha, α = 0.05

Sample standard deviation, s = 2 pounds

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 30\text{ pounds}\\H_A: \mu < 30\text{ pounds}$

We use one-tailed(left) t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{29.1 - 30}{\frac{2}{\sqrt{20}} } = -2.012$

Now,

$t_{critical} \text{ at 0.05 level of significance, 19 degree of freedom } = -1.729$

Since,

$t_{stat} < t_{critical}$

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis. Thus, there were enough evidence to conclude that the fishing line breaks with an average force of less than 30 pounds.

Option D) Yes, because the test statistic is -2.01

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a box contains 5 plain pencils and 5 pens. a second box contains 7 color pencils and 5 crayons. one item from each box is chosen

Ket [755]

First box: $\dfrac{pen}{total}=\dfrac{5}{10}=\dfrac{1}{2}$

Second box: $\dfrac{crayon}{total}=\dfrac{5}{12}$

Multiply them:

$\dfrac{1}{2}*\dfrac{5}{12}=\dfrac{5}{24}$

The probabilityis $\dfrac{5}{24}$