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3 years ago

Javier is 175% heavier than his brother. If Javier's brother weighs 80 pounds, how much does Javier weigh?

Mathematics

2 answers:

Anastaziya [24]3 years ago

7 0

If you multiply 1.75 by 80, you get Javier's weight, which is 140 pounds.

Art [367]3 years ago

4 0

175 percent of 80 is 60.

80+60=140

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3 years ago

g In a certain rural county, a public health researcher spoke with 111 residents 65-years or older, and 28 of them had obtained

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95% confidence interval for the percent of the 65-plus population that were getting the flu shot is [0.169 , 0.331].

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We are given that in a certain rural county, a public health researcher spoke with 111 residents 65-years or older, and 28 of them had obtained a flu shot.

Firstly, the Pivotal quantity for 95% confidence interval for the population proportion is given by;

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

where, $\hat p$ = sample proportion of residents 65-years or older who had obtained a flu shot = $\frac{28}{111}$ = 0.25

n = sample of residents 65-years or older = 111

p = population proportion of residents who were getting the flu shot

Here for constructing 95% confidence interval we have used One-sample z test for proportions.

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% level

of significance 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

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

= [ $0.25-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{111} } }$ , $0.25+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{111} } }$ ]

= [0.169 , 0.331]

Therefore, 95% confidence interval for the percent of the 65-plus population that were getting the flu shot is [0.169 , 0.331].

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