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dolphi86 [110]
3 years ago
6

For home economics class, Sandra has 5 cups of flour. She made 3 batches of cookies that each used 1.5 cups of flour. Write and

solve an expression to find the amount of flour Sandra has left after making the batches of cookies
Mathematics
2 answers:
stealth61 [152]3 years ago
8 0
Because one batch of cookies uses up 1.5 cups of flour, we can write this as:

batch of cookies = 1.5 cups of flour 

Because she made 3 batches we can say:

batch of cookies  = 1.5 cup of flour * 3 (because she has made three)

This equals to 3 batches of cookies using up 4.5 cups of flour.

Because at the beginning she had 5 cups and she has used 4.5 cups of flour for making the batches of cookies, she is left with 0.5 cups of flour:

 5 cups at the start - 4.5 cups used for cooking = 0.5 cups in the end.
Nataly_w [17]3 years ago
7 0
5=1.5 (3)+x
5=4.5+x
-4.5 -4.5
.5=x
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A pollster believes that 20% of the voters in a certain area favor a bond issue. If 81 voters are randomly sampled from the larg
lbvjy [14]

Answer:

0.3472 = 34.72% probability that the sampled fraction of voters favoring the bond issue will not differ from the true fraction by more than 0.02.

Step-by-step explanation:

To solve this question, we use the normal probability distribution and the central limit theorem.

Normal probability distribution:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this question:

Proportion p = 0.2, sample of n = 81. So

\mu = p = 0.2

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.2*0.8}{81}} = 0.0444

Approximate the probability that the sampled fraction of voters favoring the bond issue will not differ from the true fraction by more than 0.02.

Probability of the sampled proportion being between 0.2 - 0.02 = 0.18 and 0.2 + 0.02 = 0.22, which is the pvalue of Z when X = 0.22 subtracted by the pvalue of Z when X = 0.18. So

X = 0.22

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{0.22 - 0.2}{0.044}

Z = 0.45

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X = 0.18

Z = \frac{X - \mu}{s}

Z = \frac{0.18 - 0.2}{0.044}

Z = -0.45

Z = -0.45 has a pvalue of 0.3264

0.6736 - 0.3264 = 0.3472 = 34.72% probability that the sampled fraction of voters favoring the bond issue will not differ from the true fraction by more than 0.02.

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