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3 years ago

13/20-3/10 estimate: closer to 1 or 0?

Mathematics

2 answers:

shusha [124]3 years ago

8 0

$\frac{13}{20}\frac{12}{20}=\frac{6}{10}\\ \\\frac{6}{10}$

dexar [7]3 years ago

6 0

So, what you need to is find the decimal or percent of them. So 13/20 is 65% or .65. You get that buy multiplying it by 5 to get the denominator of 100. A denominator can go up to 100. 13/20 is closer to 1 because 65% is closer to 100% than 0%. Next 3/10 is 30%. Do that by multiplying it by 10 to get the denominator of 100. 30% is closer to 0 than 100%. Subtract them. 1-0= 1. The answer is 1.

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2 years ago

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3 years ago

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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

fomenos

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

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

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

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

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

= [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].

5 0

3 years ago

HELPPP 15 POINTS!!

yanalaym [24]

Answer:

Option C, 30

Step-by-step explanation:

Step 1: Identify how many people spent $200

There was only 30 people that spent $200

Answer: Option C, 30

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3 years ago