<span>P(3 Free Throws) = P(One Free Throw & One Free Throw & One
Free Throw)
P(3 Free Throws) = P(One Free Throw) x P(One Free Throw) x P(One Free Throw)
P(3 Free Throws) = (7/10) x (7/10) x (7/10)
P(3 Free Throws) = (7x7x7)/(10x10x10)
P(3 Free Throws) = 343/1000 </span>
<span>
P(3 Free Throws) = 0.343
The probability is 0.343 to make all 3 free throws. </span>
<span> </span>
What do you need me to answer
Answer:
Brock completed approximately 38 pull ups in the initial stage of training.
Step-by-step explanation:
We are given the following information in the question:
Ration of pull ups and box jumps initially =
Ration of pull ups and box jumps later =
Number of box jumps completed in the end of training = 53
Total number of exercises done in the end = 2x + 7x = 9x
We use the concept of ratio.
![7x = 53\\\\=x = \displaystyle\frac{53}{7}\\\\\Rightarrow 2x = \frac{106}{7}\\\\\Rightarrow \text{Total number of excercise} = \frac{477}{7}](https://tex.z-dn.net/?f=7x%20%3D%2053%5C%5C%5C%5C%3Dx%20%3D%20%5Cdisplaystyle%5Cfrac%7B53%7D%7B7%7D%5C%5C%5C%5C%5CRightarrow%202x%20%3D%20%5Cfrac%7B106%7D%7B7%7D%5C%5C%5C%5C%5CRightarrow%20%5Ctext%7BTotal%20number%20of%20excercise%7D%20%3D%20%5Cfrac%7B477%7D%7B7%7D)
Initially,
Total number of exercise = 5y + 4y
![9y = \displaystyle\frac{477}{7}\\\\\Rightarrow y = \frac{477}{63}\\\\\Rightarrow \text{Pull ups done initially} = 5y = 5\times \frac{477}{63} = 37.85 \approx 38](https://tex.z-dn.net/?f=9y%20%3D%20%5Cdisplaystyle%5Cfrac%7B477%7D%7B7%7D%5C%5C%5C%5C%5CRightarrow%20y%20%3D%20%5Cfrac%7B477%7D%7B63%7D%5C%5C%5C%5C%5CRightarrow%20%5Ctext%7BPull%20ups%20done%20initially%7D%20%3D%205y%20%3D%205%5Ctimes%20%5Cfrac%7B477%7D%7B63%7D%20%3D%2037.85%20%5Capprox%2038)
Thus, he completed approximately 38 pull ups in the initial stage of training.