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3 years ago

9

A volleyball player sets the ball in the air, and the height of the ball after t seconds is given in feet by h= -16^2+12t+6. A t

eammate wants to wait until the ball is 8 feet in the air before she spikes it. When should the teammate spike the ball? How many reasonable solutions to this problem are there?

1 answer:

JulsSmile [24]3 years ago

7 0

h=-16²+12t+6

8=-256+12t+6

256-6+8=12t

-258=12t

t=-258/12

t=21.5

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Four buses carrying 146 high school students arrive to Montreal. The buses carry, respectively, 32, 44, 28, and 42 students. One

Naily [24]

Answer:

The expected value of X is $E(X)=\frac{2754}{73} \approx 37.73$ and the variance of X is $Var(X)=\frac{226192}{5329} \approx 42.45$

The expected value of Y is $E(Y)=\frac{73}{2} \approx 36.5$ and the variance of Y is $Var(Y)=\frac{179}{4} \approx 44.75$

Step-by-step explanation:

(a) Let X be a discrete random variable with set of possible values D and probability mass function p(x). The expected value, denoted by E(X) or $\mu_x$ , is

$E(X)=\sum_{x\in D} x\cdot p(x)$

The probability mass function $p_{X}(x)$ of X is given by

$p_{X}(28)=\frac{28}{146} \\\\p_{X}(32)=\frac{32}{146} \\\\p_{X}(42)=\frac{42}{146} \\\\p_{X}(44)=\frac{44}{146}$

Since the bus driver is equally likely to drive any of the 4 buses, the probability mass function $p_{Y}(x)$ of Y is given by

$p_{Y}(28)=p_{Y}(32)=p_{Y}(42)=p_{Y}(44)=\frac{1}{4}$

The expected value of X is

$E(X)=\sum_{x\in [28,32,42,44]} x\cdot p_{X}(x)$

$E(X)=28\cdot \frac{28}{146}+32\cdot \frac{32}{146} +42\cdot \frac{42}{146} +44 \cdot \frac{44}{146}\\\\E(X)=\frac{392}{73}+\frac{512}{73}+\frac{882}{73}+\frac{968}{73}\\\\E(X)=\frac{2754}{73} \approx 37.73$

The expected value of Y is

$E(Y)=\sum_{x\in [28,32,42,44]} x\cdot p_{Y}(x)$

$E(Y)=28\cdot \frac{1}{4}+32\cdot \frac{1}{4} +42\cdot \frac{1}{4} +44 \cdot \frac{1}{4}\\\\E(Y)=146\cdot \frac{1}{4}\\\\E(Y)=\frac{73}{2} \approx 36.5$

(b) Let X have probability mass function p(x) and expected value E(X). Then the variance of X, denoted by V(X), is

$V(X)=\sum_{x\in D} (x-\mu)^2\cdot p(x)=E(X^2)-[E(X)]^2$

The variance of X is

$E(X^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{X}(x)$

$E(X^2)=28^2\cdot \frac{28}{146}+32^2\cdot \frac{32}{146} +42^2\cdot \frac{42}{146} +44^2 \cdot \frac{44}{146}\\\\E(X^2)=\frac{10976}{73}+\frac{16384}{73}+\frac{37044}{73}+\frac{42592}{73}\\\\E(X^2)=\frac{106996}{73}$

$Var(X)=E(X^2)-(E(X))^2\\\\Var(X)=\frac{106996}{73}-(\frac{2754}{73})^2\\\\Var(X)=\frac{106996}{73}-\frac{7584516}{5329}\\\\Var(X)=\frac{7810708}{5329}-\frac{7584516}{5329}\\\\Var(X)=\frac{226192}{5329} \approx 42.45$

The variance of Y is

$E(Y^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{Y}(x)$

$E(Y^2)=28^2\cdot \frac{1}{4}+32^2\cdot \frac{1}{4} +42^2\cdot \frac{1}{4} +44^2 \cdot \frac{1}{4}\\\\E(Y^2)=196+256+441+484\\\\E(Y^2)=1377$

$Var(Y)=E(Y^2)-(E(Y))^2\\\\Var(Y)=1377-(\frac{73}{2})^2\\\\Var(Y)=1377-\frac{5329}{4}\\\\Var(Y)=\frac{179}{4} \approx 44.75$

8 0

3 years ago

A project has an initial cost of $60,000, expected net cash inflows of $14,000 per year for 9 years, and a cost of capital of 14

vredina [299]

Answer:

74

Step-by-step explanation:

60,000 + 14,000

ANS 14%

ANSWER: 74

6 0

3 years ago

Solve 4x^2 - 36x + 32 = 0<br><br>Sorry! It's suppose to be + 32.<br>Please show all work.

mylen [45]

Answer:

x=1, x=8

Step-by-step explanation:

4x2−36x+32=0

Step 1: Factor left side of equation.

4(x−1)(x−8)=0

Step 2: Set factors equal to 0.

x−1=0 or x−8=0

(+1) (+8)

x=1 or x=8

6 0

3 years ago

I just wanted to see if this was 8 or negative 8 :)

nirvana33 [79]

Answer:

It's a positive 8.

Step-by-step explanation:

You can't have a negative amount of money.

4 0

3 years ago

Question 16 (Essay Worth 7 points)<br><br> Verify the identity.<br><br> tan (x + π/2) = -cot x

Rus_ich [418]

Step-by-step explanation:

We know that tan=sin/cos, so tan(x+π/2)=

$\frac{sin(x+pi/2)}{cos(x+pi/2)}$

Then, we know that sin(u+v)=sin(u)cos(v)+cos(u)sin(v),

so our equation is then

$\frac{sin(x)cos(\pi/2)+cos(x)sin(\pi/2)}{cos(x+\pi/2)} = \frac{cos(x)}{cos(x+\pi/2) }$

Then, cos(u+v)=cos(u)cos(v)-sin(u)sin(v), so our expression is then

$\frac{cos(x)}{cos(x)cos(\pi/2)-sin(x)sin(\pi/2)} = \frac{cos(x)}{-sin(x)} = -cot(x)$

6 0

3 years ago