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Elena L [17]
2 years ago
9

Phil baked two kinds of pie. Each pie pan was the same size. He served 1/2 of the blueberry pie. He served 1/4 of the apple pie.

If each pie had 8 pieces to start, what fractions in eights of the apple pie did he serve? How many more pieces of the blueberry pie than the apple pie did he serve?
Mathematics
2 answers:
USPshnik [31]2 years ago
6 0
He served 2/8 of the apple pie. And he served 2 more slices of the blueberry pie than the apple pie
suter [353]2 years ago
3 0

As per the problem,

Phil served 1/2 of the blueberry pie.

Phil served 1/4 of the apple pie.

Each pie pan was the same size.

Also each pie had 8 pieces to start.

So pieces of the apple pie served=\frac{1}{4}*8=2

So fraction of the apple pie served=\frac{2}{8}

A total of 2 pieces of apple pie were served out of 8 pieces.

Number of pieces of Blueberry pie served=\frac{1}{2}*8=4\\

Hence a total of 4 pieces of blueberry pie were served out of 8 pieces.

Hence number of pieces of blueberry pie served was 2 more than the number of pieces of apple pie served.

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Answer: 99% of confidence interval for the population proportion of employed individuals who work at home at-least once per week

//0.20113,0.20887[/tex]

Step-by-step explanation:

<u>step 1:-</u>

Given sample  size n=200

of the 200 employed individuals surveyed 41 responded that they did work at home at least once per week

 Population proportion of employed individuals who work at home at least once per week  P = \frac{x}{n} =\frac{41}{200} =0.205

Q=1-P= 1-0.205 = 0.705

<u>step 2:-</u>

Now  \sqrt{\frac{P Q}{n} } =\sqrt{\frac{(0.205)(0.705)}{200} }

=0.0015

<u>step 3:-</u>

<u>Confidence intervals</u>

<u>using formula</u>

(P  -  Z_∝} \sqrt{\frac{P Q}{n},} (P  +  Z_∝} \sqrt{\frac{P Q}{n},

(0.205-2.58(0.0015),0.205+2.58(0.0015)\\0.20113,0.20887

=0.20113,0.20887[/tex]

<u>conclusion:</u>-

99% of confidence interval for the population proportion of employed individuals who work at home at-least once per week

//0.20113,0.20887[/tex]

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