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Serga [27]
1 year ago
8

In American football, the ball is punted by dropping and kicking it before it hits the ground. The height h(t) of the football a

bove the ground in meters t seconds after being punted is affected by gravity and by the punter's kick, and can be represented as the difference of two functions: a(t)=1.4t2, which specifies the effect of gravity on the height of the ball, and v(t)=12.1t+2.5, which specifies the effect of the punt on the ball. With these two functions, h(t)=v(t)-a(t), how high above the ground was the ball when it was punted?
Mathematics
1 answer:
lakkis [162]1 year ago
4 0

Answer:

h = 8.845 m

Step-by-step explanation:

The height h(t) of the football above the ground in meters t seconds after being punted is affected by gravity and by the punters kick, and can be represented as the difference of two functions:

a(t)=1.4t^2\\\\v(t)=12.1t+2.5

We need to find h(t) such that, h(t)=v(t)-a(t)

So,

h(t)=12.1t+2.5-1.4t^2\\\\h(t)=-1.4t^2+12.1t+2.5

It is a quadratic equation. When we solve it we get :

h(t)=8.845\ m, -0.2\ m

Neglecting negative value,

h(t) = 8.845 m

So, the ball was at a height of 8.845 m when it was punted.

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Suppose that a manager is interested in estimating the average amount of money customers spend in her store. After sampling 36 t
musickatia [10]

Answer:

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

Step-by-step explanation:

Assuming this complete question: Which statement gives a valid interpretation of the interval?

The store manager is 90% confident that the average amount spent by the 36 sampled customers is between S38.01 and $44.29.

There is a 90% chance that the mean amount spent by all customers is between S38.01 and $44.29.

There is a 90% chance that a randomly selected customer will spend between S38.01 and $44.29.

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

8 0
2 years ago
One box of crackers costs $1.75. The crackers are advertised as “3 boxes for $5.25.” Which proportion can be used to represent t
Daniel [21]
A.  1/1.75 = 3/5.25 


Ignore this part I'm just trying to get at least 20 characters.
5 0
2 years ago
Read 2 more answers
Bill, George, and Ross, in order, roll a die. The first one to roll an even number wins and the game is ended. What is the proba
hram777 [196]

Answer:0.5714\frac{1}{15}

Step-by-step explanation:

Given that Bill, George, and Ross, in order, roll a die.

The first one to roll an even number wins and the game is ended.

Since Bill starts the game he can win by throwing even number or lose by throwing odd number

P(win) = 0.5, otherwise, the die will go to George.  For Bill to win, both George and Ross should throw an odd number so that Bill again gets the chance with game non ending.

Thus we have Prob of Bill winning =P of Bill winning in I throw +P of Bill winning in his II chance of throw +....infinitely

To get back the dice once he loses probability

= p both throws odd = 0.5(0.5) = 0.25

Thus Prob for Bill winning

= 0.5+0.25(0.5)^2+0.25(0.5)^3(0.25)+...

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Sum = \frac{a}{1-r} =\frac{0.5}{1-0.125} =0.5714

4 0
1 year ago
Fred buys a video game disk for $4 there was a discount of 20% what is the sales price please show work
polet [3.4K]

multiply the price by the discount then subtract the answer from the price

4 * 0.20 = 0.80

4-0.80 = 3.20

 sale price is $3.20

8 0
1 year ago
5(n+2)-8=2n<br> what s the solution for x??
PIT_PIT [208]
5n+10-8=2n
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6 0
2 years ago
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