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Illusion [34]
3 years ago
8

Three students are from a plate of cookies.Daniel ate 1/12 of the cookies, Mariel ate 4/12 of the cookies, and Tamika ate 2/12 o

f the cookies.What fraction of the cookies was left on the plate?
Mathematics
2 answers:
Mrac [35]3 years ago
8 0

Answer:

5/12  of cookies left on the plate

Step-by-step explanation:

we have to add to know what part of the dish ate and then subtract it from the total

1/12 + 4/12 + 2/12 =

when we are adding fractions with the same denominator we simply add the above

1 + 4 + 2 = 7

1/12 + 4/12 + 2/12 = 7/12

now to the total of the plate that would be 1 we subtract what they ate

1 - 7/12 =

the whole 1 can be expressed as 12/12 since if we do the 12/12 = 1 account

12/12 - 7/12

for the subtraction with the same denominator the same thing happens as for the sum, we only do it with the top

12 - 7 = 5

12/12 - 7/12 = 5/12

Rom4ik [11]3 years ago
7 0

Answer: the fraction of cookies left on the plate was 5/12.

Step-by-step explanation:

The initial amount of cookies on the plate is taken as 1.

Daniel ate 1/12 of the cookies, Mariel ate 4/12 of the cookies, and Tamika ate 2/12 of the cookies. This means that the total number of cookies that they ate is

1/12 + 4/12 + 2/12 = 7/12

Therefore, the fraction of the cookies was left on the plate would be

1 - 7/12 = (12 - 7)/12 = 5/12

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timofeeve [1]

Part a)

It was given that 3% of patients gained weight as a side effect.

This means

p = 0.03

q = 1 - 0.03 = 0.97

The mean is

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The standard deviation is

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We apply the Continuity Correction Factor(CCF)

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We convert to z-scores.

P(23.5 \: < \: X \: < \: 24.5) = P( \frac{23.5 - 19.29}{4.33} \: < \: z \: < \:  \frac{24.5 - 19.29}{4.33} ) \\  = P( 0.97\: < \: z \: < \:  1.20) \\  = 0.051

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P(X≤24)

We apply the continuity correction factor to get;

P(X<24+0.5)=P(X<24.5)

We convert to z-scores to get:

P(X \: < \: 24.5) = P(z \: < \:  \frac{24.5 - 19.29}{4.33} )  \\ =   P(z \: < \: 1.20)  \\  = 0.8849

Part c)

We want to find the probability that

11 or more patients will gain weight as a side effect.

P(X≥11)

Apply correction factor to get:

P(X>11-0.5)=P(X>10.5)

We convert to z-scores:

P(X \: > \: 10.5) = P(z \: > \:  \frac{10.5 - 19.29}{4.33} )  \\ = P(z \: > \:  - 2.03)

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Part d)

We want to find the probability that:

between 24 and 28, inclusive, will gain weight as a side effect.

P(24≤X≤28)=

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Convert to z-scores:

P(23.5  \:  <  \: X \:  <  \: 28.5) = P( \frac{23.5 - 19.29}{4.33}   \:  <  \: z \:  <  \:  \frac{28.5 - 19.29}{4.33} ) \\  = P( 0.97\:  <  \: z \:  <  \: 2.13) \\  = 0.1494

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