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Jlenok [28]
3 years ago
10

A basketball player shoots a basketball that reaches a height above 15 feet before landing back on the ground exactly after 7 se

conds. Consider the following representations. I. -(x - 3)2 + 16 II. -x2 + 8x - 7 III. -(x - 3)2 + 14 IV. -x2 + 6x + 7 Which of the representations are CORRECT for this scenario?
Mathematics
1 answer:
Papessa [141]3 years ago
3 0

Answer:

I and IV

Step-by-step explanation:

Since the height of the basketball reaches above 15 feet, hence the maximum of the function should be greater than 15 feet. Also at 7 seconds, the ball is on the ground, hence f(7) = 0 feet

The maximum of a function is at x = -b/2a

i) f(x) = -(x-3)² + 16 = -(x² - 6x + 9) + 16 = -x² + 6x + 7

The maximum of a function is at x = -b/2a = -6 / 2(-1) = 3

f(3) = -(3-3)² + 16 = 16 > 15

Also f(7) = - (7 - 3)² + 16 = 0

Hence this option is correct

ii) f(x) = -x² + 8x - 7

The maximum of a function is at x = -b/2a = -8 / 2(-1) = 4

f(4) = -4² + 8(4) - 7 = 9 < 15   not correct

Also f(7) = - 7² + 8(7) - 7 = 0

Hence this option is not correct since the maximum f(4) = 9 < 15  

iii)  f(x) = -(x-3)² + 14 = -(x² - 6x + 9) + 14 = -x² + 6x + 5

The maximum of a function is at x = -b/2a = -6 / 2(-1) = 3

f(3) = -(3-3)² + 14 = 14 < 15

Also f(7) = - (7 - 3)² + 14 = -2

Hence this option is not correct since the maximum f(4) = 9 < 15 and f(7) ≠ 0

iv)f(x)  = -x² + 6x + 7

The maximum of a function is at x = -b/2a = -6 / 2(-1) = 3

f(3) = -(3)² + 6(3) + 7 = 16 > 15

Also f(7) = - (7)² + 6(7) + 7 = 0

Hence this option is correct

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Answer:

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The probability distribution of <em>X</em> id provided.

The formula to compute the mean or expected value of <em>X </em>is:

\mu=E(X)=\sum x.P(X=x)

The formula to compute the standard deviation of <em>X </em>is:

\sigma=\sqrt{E(X^{2})-(E(X))^{2}}

The formula of E (X²) is:

E(X^{2})=\sum x^{2}.P(X=x)

(a)

Compute the expected value of <em>X</em> as follows:

E(X)=\sum x.P(X=x)\\=(0\times0.06)+(1\times0.25)+(2\times0.35)+(3\times0.15)+(4\times0.13)+(5\times0.06)\\=2.22\\\approx2.2

Thus, the mean or expected value of <em>X </em>is 2.2.

(b)

Compute the value of E (X²) as follows:

E(X^{2})=\sum x^{2}.P(X=x)\\=(0^{2}\times0.06)+(1^{2}\times0.25)+(2^{2}\times0.35)+(3^{2}\times0.15)+(4^{2}\times0.13)+(5^{2}\times0.06)\\=6.58

Compute the standard deviation of <em>X</em> as follows:

\sigma=\sqrt{E(X^{2})-(E(X))^{2}}\\=\sqrt{6.58-(2.22)^{2}}\\=\sqrt{1.6516}\\=1.285\\\approx1.3

Thus, the standard deviation of <em>X</em> is 1.3.

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