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marshall27 [118]

3 years ago

10

Wayne is at a basketball practice preparing to shoot three pointers. The probability of wine making a three pointer is 21%. What

is the probability he makes his first three pointer on the fourth shot.
A. 8%
B. 10%
C. 15%
D. 17%

2 answers:

nlexa [21]3 years ago

8 0

Answer:

the correct answer is B. 10%

Step-by-step explanation:

<em>From the given question, we recall the following,</em>

<em>The probability of Wayne making a three pointer = 21%</em>

<em>Now, what is the the probability he makes his first three pointer on the fourth shot.</em>

<em>The next step is to find a solution to the question</em>

<em>The probability of not shooting a 3 pointer is 1- 0.79 = 1-p=q</em>

<em>All the throws of wayne are independent of each other</em>

<em>If two or ore event are independent we have,</em>

<em>P (AnB) = P(A) * P(B)</em>

<em>P (AnBnC) = P(A) *P (B) *P (C)</em>

<em>P (AnBnCnD) = P(A) * P(B)* P(C) *P(D)</em>

<em>Now, let make first 3 pointer on the 4th shot, we have:</em>

<em>q³= p (0.79)³ (0.21) =10% or (1-0.21) x (1-0.21)x (1-0.21) x 0.21=(0.79) x (0.79) x (0.79) x(0.21)=0.1</em>

<em>which is =10%</em>

patriot [66]3 years ago

5 0

Answer:

The answer is A. Was this a test question or a normal lesson?

Step-by-step explanation:

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Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

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P.Q. = $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t_n__1+n_2-2$

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$n_1$ = Sample size for Brand B data = 13

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$s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} }$ = 3.013

Here, $s^{2}_X$ and $s^{2} _Y$ are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < $t_2_0$ < 2.528) = 0.98

P(-2.528 < $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 2.528) = 0.98

P(-2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(Xbar -Ybar) -(\mu_X-\mu_Y)$ < 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

P( (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(\mu_X-\mu_Y)$ < (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ , (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ]

[ $(36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ , $(36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

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