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cestrela7 [59]
3 years ago
7

Francesca had 20 dollars to spend on 3 gifts. She spent 8 1 4 dollars on gift A and 4 4 5 dollars on gift B. How much money did

she have left for gift C? Solve on paper to find the answer as a fraction. Then check your work on Zearn! Francesca had dollars for gift C.
Mathematics
1 answer:
Karo-lina-s [1.5K]3 years ago
8 0

Answer:

Francesca had \$\frac{139}{20} \ \ OR \ \ \$6\frac{19}{20} money left for gift C.

Step-by-step explanation:

Given:

Total money to spent on gift = $20

Money Spent on gift A = \$ 8\frac{1}{4}

\$ 8\frac{1}{4} can Rewritten as  \$\frac{33}{4}

Money Spent on gift A = \$\frac{33}{4}

Money Spent on gift B = \$ 4\frac{4}{5}

\$ 4\frac{4}{5} can Rewritten as  \$\frac{24}{5}

Money Spent on gift B = \$\frac{24}{5}

We need To find the Money left for gift C.

Solution:

Money left for gift C can be calculated by subtracting Money Spent on gift A and Money Spent on gift B with Total Money spent on gifts.

Framing in equation form we get;

Money left for gift C = 20-\frac{33}{4}-\frac{24}{5}

Now taking LCM for making the denominator common we get;

Money left for gift C =\frac{20\times 20}{20}-\frac{33\times5}{4\times5}-\frac{24\times4}{5\times4}= \frac{400}{20}-\frac{165}{20}-\frac{96}{20}=\frac{400-165-96}{20} = \$\frac{139}{20} \ \ OR \ \ \$6\frac{19}{20}

Hence Francesca had \$\frac{139}{20} \ \ OR \ \ \$6\frac{19}{20} money left for gift C.

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You have drawn a simple random sample of 36 college students, asking each student how much rent they pay per month. You obtain a
Karo-lina-s [1.5K]

Answer:

573-1.690\frac{22}{\sqrt{36}}=566.8033    

573+1.690\frac{22}{\sqrt{36}}=579.1967      

And the 90% confidence interval would be between [566.80 and 579.20]

Step-by-step explanation:

Information given

\bar X=573 represent the sample mean

\mu population mean (variable of interest)

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Confidence interval

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

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The degrees of freedom are given by:

df=n-1=36-1=35

The Confidence level is 0.90 or 90%, the significance would be \alpha=0.1 and \alpha/2 =0.05, and the critical value for this case would be t_{\alpha/2}=1.690

And replacing we got:

573-1.690\frac{22}{\sqrt{36}}=566.8033    

573+1.690\frac{22}{\sqrt{36}}=579.1967    

And the 90% confidence interval would be between [566.8033 and 579.1967]

6 0
3 years ago
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3 0
3 years ago
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