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3 years ago

The local ice cream parlor sells an average of 11 milkshakes a day for $3.99

Mathematics

1 answer:

Greeley [361]3 years ago

6 0

In a week, the parlor would make about $300 from milkshakes

The total amount the ice cream parlor makes in a week is dependent is the product of the number of days in a week , the number of milkshakes sold in a day and the price of a milkshake.

Total amount = number of milkshakes sold in a week x price of a milkshake

There are 7 days in a week, in a week the parlor would sell : 11 x 7 = 77 milkshakes

price of a milkshake = $3.99

77 x $3.99 = $307.23

A similar question was solved here: brainly.com/question/14090575?referrer=searchResults

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3 years ago

At Western University the historical mean of scholarship examination scores for freshman applications is 900. A historical popul

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Answer:

a) Null Hypothesis: $\mu =900$

Alternative hypothesis: $\mu \neq 900$

b) The 95% confidence interval would be given by (910.05;959.95)

c) Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

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a. State the hypotheses.

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b. What is the 95% confidence interval estimate of the population mean examination score if a sample of 200 applications provided a sample mean x¯¯¯= 935?

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$\bar X=935$ represent the sample mean for the sample

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$\sigma=180$ represent the population standard deviation

n=200 represent the sample size

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=3278.222$

The sample deviation calculated $s=97.054$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$