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Triss [41]
3 years ago
9

The hang time of a basketball player who jumps a verticle distance of 2 feet (0.6m) is 2/3 seconds. What will be the hang time i

f the player reaches the same height while jumping a horizontal distance of 4 feet (1.2m) ?
Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
8 0
The hang time will be the same because the amount of time spent in the air is dependent on the vertical height. The fact that the player covers horizontal distance is simply due to him also pushing forward, instead of only upwards.
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Daca a/3=7/10 calculati 100a
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A/3=7/10
21=10a
a=2.1
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3 years ago
HELP ASAP!!!!!!!! PLSSSSSSS HURRY!!!
tamaranim1 [39]

Answer:

D

Step-by-step explanation:

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3 years ago
The Lacrosse booster club is holding a raffle for a fundraiser. They will sell 100 tickets for $5 each and select 4 winners. All
Nadusha1986 [10]

Answer:

<h2>There are 3,921,225 ways to select the winners.</h2>

Step-by-step explanation:

This problem is about combinations with no repetitions, because the same person can't win four times. It's a combinaction because the order of winning doesn't really matter.

Combinations without repetitions are defined as

C_{n}^{r}  =\frac{n!}{r!(n-r)!}

Where n=100 and r=4.

Replacing values, we have

C_{100}^{4}  =\frac{100!}{4!(100-4)!}=\frac{100!}{4! 96!}=\frac{100 \times 99 \times 98 \times 97 \times 96!}{4! \times 96!}=  \frac{94,109,400}{24}= 3,921,225

Therefore, there are 3,921,225 ways to select the winners.

Additionally, as you can imagine, the probability of winning is extremely low, it would be 3,921,225 to 1.

3 0
3 years ago
PLEASE HELP!!!! see attachment below… I bet you cant solve this:
d1i1m1o1n [39]

Answer:

  • 0

Step-by-step explanation:

<u>For each odd i the term is:</u>

  • (-1)^{i} (5!*5/4!)^{1/2} = - \sqrt{25}

<u>For each even i the term is:</u>

  • (-1)^{i} (5!*5/4!)^{1/2} =  \sqrt{25}

So the sum of the first 100 terms is zero

3 0
2 years ago
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