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geniusboy [140]
2 years ago
5

How many quarts of water must be added to 2 gal of soup that is 50% chicken broth to make the soup 40% chicken broth?

Mathematics
1 answer:
timofeeve [1]2 years ago
7 0

Step-by-step explanation: Well we would know that the original soup is one gallon water and one gallon broth because that is 2 gallons 50% to 50%.

Then to decrease the amount of broth by 10%, we would accomplish this by first turning the original gallons into quarts so we would have 4 quarts broth and 4 quarts water. We need to get a total of ten quarts SO........ we know that we need to add 2 MORE quarts of WATER.

Finally when we add all the quarts together we get 10 total with 6 water and 4 broth which totals to 60% water and 40% broth.

                           FINAL ANSWER: 2 QUARTS WATER

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Answer:

a) The 95% confidence interval would be given (0.0702;0.1098).

We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)

b) Since the confidence interval NOT contains the value 0.06 we  have anough evidence to reject the claim at 5% of significance.  

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

p represent the real population proportion of interest  

\hat p =0.09 represent the estimated proportion for the sample  

n=800 is the sample size required  

z represent the critical value for the margin of error  

Confidence =0.95 or 95%

Part a

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

The confidence interval would be given by this formula  

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.96  

The margin of error is given by :

Me=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

Me=1.96 \sqrt{\frac{0.09(1-0.09)}{800}}=0.0198

And replacing into the confidence interval formula we got:  

0.09 - 1.96 \sqrt{\frac{0.09(1-0.09)}{800}}=0.0702  

0.09 + 1.96 \sqrt{\frac{0.108(1-0.09)}{800}}=0.1098  

And the 95% confidence interval would be given (0.0702;0.1098).

We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)

Part b

Since the confidence interval NOT contains the value 0.06 we  have anough evidence to reject the claim at 5% of significance.

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Step-by-step explanation:

If a triangle is rotated 90° about the origin.

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The correct statement is:
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Answer:

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