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erma4kov [3.2K]

3 years ago

9

A volleyball reaches its maximum height of 13 feet, 3 seconds after its served. Which of the following quadratics could model th

e height of the vollyball over time after it is served. Select all that apply.
A: f(x)=2x^2+12x+5
B: f(x)=-2x^2+12x-5
C: f(x)=-2x^2-12x+5
D: f(x)=-2(x-3)^2+13
E: f(x)=-2(x+3)^2+13

2 answers:

Ivahew [28]3 years ago

7 0

Answer:

B: f(x)=-2x^2+12x-5

and!!!!!!

D: f(x)=-2(x-3)^2+13

Step-by-step explanation:

b and are the only ones that give you the values 13 and 3

kykrilka [37]3 years ago

3 0

If it reaches its maximum height a t=3 its velocity must be zero at t=3

dy/dx=0 at t=3

dA/dx=4x+12=24 at t=3 so this does not apply

dB/dx=-4x+12=0 at t=3 and B=13 at t=3 so B applies

dC/dx=-4x-12=-24 at t=3 so this does not apply

dD/dx=-4x-12=-24 at t=3 so this does not apply

dE/dx=-4x-12=-24 at t=3 so this does not apply

So only quadratic B: reaches a maximum height of 13 feet at 3 seconds.

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Suppose approximately 75% of all marketing personnel are extroverts, whereas about 70% of all computer programmers are introvert

Setler [38]

Answer:

$P(x \ge 5) = 1.000$ ---- At least 5 from marketing departments are extroverts

$P(x=15) = 0.013$ ---- All from marketing departments are extroverts

$P(x = 0) = 0.002$ ---------- None from computer programmers are introverts

Step-by-step explanation:

See comment for complete question

The question is an illustration of binomial probability where

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$(a):\ P(x \ge 5)$

$n = 15$ --- marketing personnel

$p = 75\%$ --- proportion that are extroverts

Using the complement rule, we have:

$P(x \ge 5) = 1 - P(x < 5)$

So, we have:

$P(x < 5) =P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)$

$P(x = 0) = ^{15}C_{0} * (75\%)^0 * (1 - 75\%)^{15 - 0} = 1 * 1 * (0.25)^{15} = 9.31 * 10^{-10}$

$P(x = 1) = ^{15}C_{1} * (75\%)^1 * (1 - 75\%)^{15 - 1} = 15* (0.75)^1 * (0.25)^{14} = 4.19 * 10^{-8}$

$P(x = 2) = ^{15}C_{2} * (75\%)^2 * (1 - 75\%)^{15 - 2} = 105* (0.75)^2 * (0.25)^{13} = 8.80 * 10^{-7}$

$P(x = 3) = ^{15}C_{3} * (75\%)^3 * (1 - 75\%)^{15 - 3} = 455* (0.75)^2 * (0.25)^{12} = 0.0000153$

$P(x = 4) = ^{15}C_{4} * (75\%)^4 * (1 - 75\%)^{15 - 4} = 1365 * (0.75)^4 * (0.25)^{11} = 0.000103$

So, we have:

$P(x < 5) = (9.31 * 10^{-10}) + (4.19 * 10^{-8}) + (8.80 * 10^{-7}) + 0.0000153 + 0.000103$

$P(x < 5) = 0.00011922283$

Recall that:

$P(x \ge 5) = 1 - P(x < 5)$

$P(x \ge 5) = 1 - 0.00011922283$

$P(x \ge 5) = 0.9998$

$P(x \ge 5) = 1.000$ --- approximated

$(b)\ P(x = 15)$

$n = 15$ --- marketing personnel

$p = 75\%$ --- proportion that are extroverts

So, we have:

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$P(x=15) = ^{15}C_{15} * 0.75^{15} * (1 - 0.75)^{15-15}$

$P(x=15) = 1 * 0.75^{15} * (0.25)^{0$

$P(x=15) = 0.013$

$(c)\ P(x = 0)$

$n=5$ ---------- computer programmers

$p = 70\%$ --- proportion that are introverts

So, we have:

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$P(x = 0) = ^{5}C_0 * (70\%)^0 * (1 - 70\%)^{5-0}$

$P(x = 0) = 1 * 1 * (0.30)^5$

$P(x = 0) = 0.002$

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