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mojhsa [17]
1 year ago
6

in a basketball game tatiana made 23 baskets . each of the baskets was worth either 2 or 3 points and Tatiana scored a total of

53 points . let x represent the number of two point baskets she made and y represent the number of 3 point baskets she made

Mathematics
2 answers:
natima [27]1 year ago
8 0
Here is the answer to you question!

Volgvan1 year ago
5 0

Answer:

<h2>n a basketball game tatiana made 23 baskets . each of the baskets was worth either 2 or 3 points and Tatiana scored a total of 53 points . let x represent the number of two point baskets she made and y represent the number of 3 point baskets she made</h2>

Step-by-step explanation:

We know that Tatiana made 23 baskets in a game.

x represents two-points baskets, and y represents three-points baskets.

If she scored 53 points, we would have the following relations

x+y=23, which represents the total number of baskets.

2x+3y=53, which represents the total number of points.

To solve this system, we multiply the first equation by -2, then sum and solve for y

\left \{ {{-2x-2y=-46} \atop {2x+3y=53}} \right.\\ y=7

Then, we replace this value to find the other one

x+y=23\\x+7=23\\x=23-7\\x=16

Therefore, Tatiana made 7 baskets of two points and 16 baskets of three points.

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(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

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We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

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where, \hat p = sample proportion of Americans who decide to not go to college = 48%

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           p = population proportion of Americans who decide to not go to

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<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>

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<u>So, 90% confidence interval for the population proportion, p is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5% level

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P(-1.645 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.645) = 0.90

P( -1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < \hat p-p < 1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.90

P( \hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.90

<u>90% confidence interval for p</u> = [ \hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ]

 = [ 0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } } , 0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } } ]

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               n = 54.79^{2}

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