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Fudgin [204]
3 years ago
12

Marshall divides his money into four categories. He saves 1/3 of his money. He gives 1/6 of his money to a charity. He uses 1/14

of his money to buy snacks at school, and he spends 3/7 of his money on fun. What fraction of his money does he save and spend on fun?
Mathematics
2 answers:
Len [333]3 years ago
7 0
1/3 + 3/7
save + fun + extra

(7/21 + 9/21) = 16/21 on save and fun
frutty [35]3 years ago
4 0

Marshall divides his money into four categories.

He saves 1/3 of his money.

He gives 1/6 of his money to a charity.

He uses 1/14 of his money to buy snacks at school,

and he spends 3/7 of his money on fun.

Money he saves = \frac{1}{3}

Money he spends for fun = \frac{3}{7}

Fraction of his money he save and spend on fun = \frac{1}{3} + \frac{3}{7}

LCD of (3 and 7) is 21

\frac{1*7}{3*7} + \frac{3*3}{7*3}

\frac{7}{21} + \frac{9}{21}

Denominators are same so we add the numerators

\frac{16}{21}

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⦁ In a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football. Construct a 95% confiden
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Answer:

95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

Step-by-step explanation:

We are given that in a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football.

Firstly, the pivotal quantity for 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is given by;

        P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } ~ N(0,1)

where, \hat p = proportion of adults in the United States whose favorite sport to watch is football in a sample of 1219 adults = \frac{354}{1219}

           n = sample of US adults  = 1291

           p = population proportion of adults

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5%

                                                    significance level are -1.96 & 1.96}

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } ) = 0.95

<u>95% confidence interval for p</u> = [ \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } ]

                         = [ \frac{354}{1219}-1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} } , \frac{354}{1219}+1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} } ]

                         = [0.265 , 0.316]

Therefore, 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

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