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ivanzaharov [21]

2 years ago

12

Gina is making bags of trail mix for hiking club. She will use 20 ounces of walnuts, 11.2 ounces of almonds, and 28.3 ounces of

cashews. This amount makes 25 bags of trail mix. How many ounces are in each bag?

2 answers:

Romashka-Z-Leto [24]2 years ago

5 0

Answer:

there will be 2.38 ounces in one bag of the trail mix

o-na [289]2 years ago

3 0

Answer:

2.15 ounces

Step-by-step explanation:

The total ounces of nuts in the bag is calculated as:

20 ounces of walnuts + 11.1 ounces of almonds + 24.8 ounces of cashews = 55.9 ounces.

From the question we are told that these amounts make 26 bags.

The number of ounces that are in each bag is calculated as:

26 bags = 55.9 ounces

1 bag = x ounces

Cross Multiply

26 bags × x ounces = 1 bag × 55.9 ounces

x ounces = 1 bag × 55.9 ounces/26 bags

x ounces = 2.15 ounces

Therefore, there are 1.94 ounces in each bag

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Consider the computer output below. Fill in the missing information. Round your answers to two decimal places (e.g. 98.76). Test

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Answer:

$SE_{Mean}=\frac{s}{\sqrt{n}}=\frac{4.77}{\sqrt{19}}=1.094$

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a) $df=n-1= 19-1= 18$

b) $p_v =2*P(t_{18}$

If we compare the p value and a significance level for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

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Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

$\bar X=98.77$ represent the sample mean

$s=4.77$ represent the sample standard deviation

n=19 represent the sample selected

$\alpha$ significance level

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Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu \neq 100$

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we can calculate the Standard error for the mean like this:

$SE_{Mean}=\frac{s}{\sqrt{n}}=\frac{4.77}{\sqrt{19}}=1.094$

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Now we have everything in order to replace into formula (1):

$98.77-1.96\frac{4.77}{\sqrt{19}}=96.625$

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Part a

The degree of freedom are given by:

$df=n-1= 19-1= 18$

Part b

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$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

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$t=\frac{98.77-100}{\frac{4.77}{\sqrt{19}}}=-1.124$

Then since is a two sided test the p value would be:

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Alternative hypothesis: $\mu > 100$

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