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MrRa [10]
3 years ago
5

A baker needs a total of 13(1/3) cups of flour to make four batches of cookies. How much flour, in cups, does she need for each

batch of cookies? Write and solve an equation to find the answer.
Mathematics
2 answers:
alexandr402 [8]3 years ago
4 0

Answer:

She needs 3\frac{1}{3} cups of flour for each batch of cookies.

Step-by-step explanation:

A baker needs a total 13\frac{1}{3} cups of flour to make four batches of cookies.

We have to calculate how much flour she needs for each batch of cookies.

Let the flour needed for one batch be f, so the expression would be

f × 4 = 13\frac{1}{3}

First we convert mixed fraction to proper fraction.

f × 4 = \frac{40}{3}

f = \frac{\frac{40}{3} }{4}

= \frac{40}{3} × \frac{1}{4} = \frac{10}{3}

Now we convert this proper fraction to mixed fraction.

\frac{10}{3} = 3\frac{1}{3}

She needs 3\frac{1}{3} cups of flour for each batch of cookies.

qaws [65]3 years ago
4 0

Answer:

she needs 3 and one-third B: trust me i took the test

Step-by-step explanation:

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Ede4ka [16]

Answer:

Explained below.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 \mu_{\hat p}= p

The standard deviation of this sampling distribution of sample proportion is:

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(a)

The sample selected is of size <em>n</em> = 450 > 30.

Then according to the central limit theorem the sampling distribution of sample proportion is normally distributed.

The mean and standard deviation are:

\mu_{\hat p}=p=0.75\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{450}}=0.0204

So, the sampling distribution of sample proportion is \hat p\sim N(0.75,0.0204^{2}).

(b)

Compute the probability that the sample proportion will be within 0.04 of the population proportion as follows:

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Thus, the probability that the sample proportion will be within 0.04 of the population proportion is 0.95.

(c)

The sample selected is of size <em>n</em> = 200 > 30.

Then according to the central limit theorem the sampling distribution of sample proportion is normally distributed.

The mean and standard deviation are:

\mu_{\hat p}=p=0.75\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{200}}=0.0306

So, the sampling distribution of sample proportion is \hat p\sim N(0.75,0.0306^{2}).

(d)

Compute the probability that the sample proportion will be within 0.04 of the population proportion as follows:

P(p-0.04

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Thus, the probability that the sample proportion will be within 0.04 of the population proportion is 0.81.

(e)

The probability that the sample proportion will be within 0.04 of the population proportion if the sample size is 450 is 0.95.

And the probability that the sample proportion will be within 0.04 of the population proportion if the sample size is 200 is 0.81.

So, there is a gain in precision on increasing the sample size.

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Step-by-step explanation:

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