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lord [1]
2 years ago
6

The students in Mr. Wilson's Physics class are making golf ball catapults. The

Mathematics
1 answer:
Mnenie [13.5K]2 years ago
8 0

Answer:

Part a) About 48.6 feet

Part b) About 8.3 feet

Part c) The domain is 0 \leq x \leq 48.6\ ft and the range is 0 \leq y \leq 8.3\ ft

Step-by-step explanation:

we have

y=-0.014x^{2} +0.68x

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

where

x is the ball's  distance from the catapult in feet

y is the flight of the balls in feet

Part a)  How far did the ball fly?

Find the x-intercepts or the roots of the quadratic equation

Remember that

The x-intercept is the value of x when the value of y is equal to zero

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

-0.014x^{2} +0.68x=0

so

a=-0.014\\b=0.68\\c=0

substitute in the formula

x=\frac{-0.68(+/-)\sqrt{0.68^{2}-4(-0.014)(0)}} {2(-0.014)}

x=\frac{-0.68(+/-)0.68} {(-0.028)}

x=\frac{-0.68(+)0.68} {(-0.028)}=0

x=\frac{-0.68(-)0.68} {(-0.028)}=48.6\ ft

therefore

The ball flew about 48.6 feet

Part b) How high above the ground did the ball fly?

Find the maximum (vertex)

y=-0.014x^{2} +0.68x

Find out the derivative and equate to zero

0=-0.028x +0.68

Solve for x

0.028x=0.68

x=24.3

<em>Alternative method</em>

To determine the x-coordinate of the vertex, find out the midpoint  between the x-intercepts

x=(0+48.6)/2=24.3\ ft

To determine the y-coordinate of the vertex substitute the value of x in the quadratic equation and solve for y

y=-0.014(24.3)^{2} +0.68(24.3)

y=8.3\ ft

the vertex is the point (24.3,8.3)

therefore

The ball flew above the ground about 8.3 feet

Part c) What is a reasonable domain and range for this function?

we know that

A  reasonable domain is the distance between the two x-intercepts

so

0 \leq x \leq 48.6\ ft

All real numbers greater than or equal to 0 feet and less than or equal to 48.6 feet

A  reasonable range is all real numbers greater than or equal to zero and less than or equal to the y-coordinate of the vertex

so

we have the interval -----> [0,8.3]

0 \leq y \leq 8.3\ ft

All real numbers greater than or equal to 0 feet and less than or equal to 8.3 feet

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Step-by-step explanation:

<u>Question 1</u>

The probability distribution table for the average price of the textbook sales in each of these three time periods is given below:

\left|\begin{array}{c|c|c|c}&$Start-of-term&$End-of-term&$Other\\$Sales proportion&0.45&0.31&0.24\\$Average price&\$82.52&\$49.12&\$65.23\end{array}\right|

We are required to calculate the average price of the textbook over all seasons.

Expected Value

= (0.45 X 82.52)+(0.31 X 49.12) +(0.24 X 65.23)\\=68.02

The average price of the textbook over all seasons is $68.02

<u>Question 2</u>

Distribution Table for Number of Courses being taken by BHCC Students

\left|\begin{array}{c|ccccc}x$(No of classes)& 1&2&3 &4&5\\P(x)&0.24&0.26&0.12&0.3 &0.08\end{array}\right|

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P(x\geq 2)=0.26+0.12+0.3+0.08=0.76

b.) Probability that a student is taking at least 3 classes

P(x\geq 3)=0.12+0.3+0.08=0.5

c.) Probability that a student is taking more than 3 classes

P(x> 3)=0.3+0.08=0.38

d.) Probability that a student is taking less than 2 classes

P(x< 2)=0.24

e.) Probability that a student is taking no more than 2 classes

P(x\leq  2)=0.24+0.26=0.5

f)Average (mean) amount of classes

=(1*0.24)+(2*0.26)+(3*0.12)+(4*0.3)+(5*0.08)\\\mu=2.72

g)Standard deviation for the amount of classes

\left|\begin{array}{c|ccccc|c}x$(No of classes)& 1&2&3 &4&5&Sum\\x-\mu&-1.72&-0.72&0.28&1.28&2.28\\(x-\mu)^2&2.9584&0.5184&0.0784&1.6384&5.1984\\P(x)&0.24&0.26&0.12&0.3 &0.08\\--&--&---&---&--&--&--\\(x-\mu)^2P(x)&0.71&0.13&0.01&0.49&0.42&1.76\end{array}\right|

Standard Deviation

=\sqrt{\sum(x-\mu)^2P(x)} \\=\sqrt{1.76} \\\sigma =1.33

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