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Luden [163]
1 year ago
15

Alisa buys a watermelon every year. The first time she bought a watermelon, it cost $2.50. She notices the price is getting more

expensive, at a rate of 3% per year.
Alisa sketches a graph of the situation, labeling the y-intercept and the point representing the price in Year 4.

Which statements are true?

Select each correct answer.

A. The graph has an x-intercept and a y-intercept.
B. The point (4, 2.81) is on the graph.
C. The graph has a y-intercept only.
D. The graph decreases from left to right.
E. The graph increases from left to right.
F. The point ​ (4, 2.62) is on the graph.
Mathematics
1 answer:
timofeeve [1]1 year ago
7 0

So the cost of water melon at year 0 is $ 2.50 since it will increase by 3% per yer

0              2.5

1              2.575

2              2.65225

3              2.7318175

4              2.813772025

So the correct statement is B. The point (4, 2.81) is on the graph.

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Peter decided to buy a new car. He made a $2,160 down payment and then took a 48-month loan. The total amount for the car, plus
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16560-2160=14400, 14400=48x, x=$300
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Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be
Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of t=0.6482

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

h(t)=-16t^2 +v_{o}t+s       Eqn(1).

Where:

h(t) : is the height range as a function of time

v_{o}   : is the initial velocity

s     : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

h(t)=-16t^2 + v_{o}t + 40        Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

<u>Ben:</u>

We know that v_{o}=60ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+60t+40        Eqn.(3)

<u>Josh:</u>

We know that v_{o}=48ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+48t+40       Eqn. (4).

<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em>h(t)=0<em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

Thus we have for Ben, Eqn. (3) gives:

h(t)=0-16t^2+60t+40=0

Using the quadratic expression to find the roots of the quadratic we have:

t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec

Since time can only be positive we reject the t_{2} solution and we keep that Ben's book took t=4.3276 seconds to reach the ground.

Similarly solving for Josh we obtain

t_{1}=3.6794sec\\t_{2}=-0.6794sec

Thus again we reject the negative and keep the positive solution, so Josh's book took t=3.6794 seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482

So to answer the original question:

<em>Whose textbook reaches the ground first and by how many seconds?</em>

  • Josh's textbook reached the ground first
  • Josh's textbook reached the ground first by a difference of t=0.6482

3 0
1 year ago
A linear regression equation is calculated for a sample of n = 20 pairs of X and Y values.
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Answer: B) 18

Step-by-step explanation:

The degree of freedom for Standard error or df(Residual) is the sample size minus the number of estimated parameters ,.

df= n - p  , where n= sample size , p = number of parameters.

According to the given problem , we have

Sample size : n=20

Number of parameters (x and y ): p= 2

Then, the df value for the standard error of estimate will be :

df= n-p =20-2=18

Thus , the df value for the standard error of estimate is 18 .

Hence, the correct answer is B) 18 .

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1. lines j and k because they have the same slope.
2.d because -1/3 is the opposite slope of 3 and it passes through that point.
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