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Lubov Fominskaja [6]
2 years ago
6

Alice has a total of 12 dimes and nickels. She has 2 more nickels than dimes. Which equation represents the given problem situat

ion?
A. c + (c + 2) = 12, where c is the number of dimes

B. c + 2c = 12, where c is the number of nickels

C. c + (c + 2) = 12, where c is the number of nickels

D. c + 2c = 12, where c is the number of dimes
Mathematics
2 answers:
const2013 [10]2 years ago
6 0

Answer:  The correct option is

(A) c+(c+2)=12, where c is the number of dimes.

Step-by-step explanation:  Given that Alice has a total of 12 dimes and nickels and she has 2 more nickels than dimes.

We are to select the correct equation that represents the given problem situation.

Let c represents the number of dimes. Then, the number of nickels will be (c + 2).

Since there are total 12 coins, so the required equation is given by

c+(c+2)=12.

Thus, the required equation is

c+(c+2)=12, where c is the number of dimes.

Option (A) is CORRECT.

Westkost [7]2 years ago
5 0

Answer:

A.

Step-by-step explanation:

She has 2 more nickels then dimes not 2 times more therefore answers B and D are incorrect. C is incorrect because it has that there are 2 more dimes than nickels. A is correct because it says that there are c dimes, and then c +2 nickels.

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Of the 500 sample households in the previous exercise, 7 had three or more large-screen TVs. (a) The percentage of households in
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Answer:

a) The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

\hat p=\frac{7}{500}=0.014

And that represent the 1.4%.

b) And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

Step-by-step explanation:

Data given and notation  

n=500 represent the random sample taken    

X=7 represent the households with three or more large-screen TVs

\hat p=\frac{7}{500}=0.014 estimated proportion of households with three or more large-screen TVs

\alpha=0.05 represent the significance level (no given, but is assumed)    

z would represent the statistic (variable of interest)    

p= population proportion of households with three or more large-screen TVs

Part a

The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

\hat p=\frac{7}{500}=0.014

And that represent the 1.4%.

Part b

Yes is possible. We hav that np>10 and n(1-p)>10 so we have the assumption of normality to find the interval.

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

And replacing into the confidence interval formula we got:

0.014 - 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.00370

0.014 + 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.0243

And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

4 0
2 years ago
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