Answer:
Step-by-step explanation:
<u>Proportional relationship equation:</u>
- y = kx, where k- is the constant of proportionality
<u>Use one pair of coordinates to find the value of k, the point (3, 2):</u>
Answer:
7
Step-by-step explanation:
Since, Sophia had three pages of homework and she only has three left you can set up an equation to solve. The equation would be 10=x-3, then you just solve the equation which would be 7, hope that helped.
Answer:
52.56% probability that eight or more of the flights will arrive on time.
Step-by-step explanation:
For each flight, there are only two possible outcomes. Either it is on time, or it is not. The probability of a flight being on time is independent from other flights. So we use the binomial probability distribution to solve this question.]
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And p is the probability of X happening.
At a certain airport, 75% of the flights arrive on time.
This means that ![p = 0.75](https://tex.z-dn.net/?f=p%20%3D%200.75)
A sample of 10 flights is studied.
This means that ![n = 10](https://tex.z-dn.net/?f=n%20%3D%2010)
Find the probability that eight or more of the flights will arrive on time.
![P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%208%29%20%3D%20P%28X%20%3D%208%29%20%2B%20P%28X%20%3D%209%29%20%2B%20P%28X%20%3D%2010%29)
In which
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 8) = C_{10,8}.(0.75)^{8}.(0.25)^{2} = 0.2816](https://tex.z-dn.net/?f=P%28X%20%3D%208%29%20%3D%20C_%7B10%2C8%7D.%280.75%29%5E%7B8%7D.%280.25%29%5E%7B2%7D%20%3D%200.2816)
![P(X = 9) = C_{10,9}.(0.75)^{9}.(0.25)^{1} = 0.1877](https://tex.z-dn.net/?f=P%28X%20%3D%209%29%20%3D%20C_%7B10%2C9%7D.%280.75%29%5E%7B9%7D.%280.25%29%5E%7B1%7D%20%3D%200.1877)
![P(X = 10) = C_{10,10}.(0.75)^{10}.(0.25)^{0} = 0.0563](https://tex.z-dn.net/?f=P%28X%20%3D%2010%29%20%3D%20C_%7B10%2C10%7D.%280.75%29%5E%7B10%7D.%280.25%29%5E%7B0%7D%20%3D%200.0563)
![P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.2816 + 0.1877 + 0.0563 = 0.5256](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%208%29%20%3D%20P%28X%20%3D%208%29%20%2B%20P%28X%20%3D%209%29%20%2B%20P%28X%20%3D%2010%29%20%3D%200.2816%20%2B%200.1877%20%2B%200.0563%20%3D%200.5256)
52.56% probability that eight or more of the flights will arrive on time.
Answer:
I tried, can you explain more please i dont get it and cant solve it
Step-by-step explanation:
Answer:
Option B is correct
![5(8) \div 5+ 3(8)](https://tex.z-dn.net/?f=5%288%29%20%5Cdiv%205%2B%203%288%29)
Step-by-step explanation:
Given the equation: ![40 \div 5+3 \cdot 8](https://tex.z-dn.net/?f=40%20%5Cdiv%205%2B3%20%5Ccdot%208)
we can write
40 as ![5 \cdot (8) = 5(8)](https://tex.z-dn.net/?f=5%20%5Ccdot%20%288%29%20%3D%205%288%29)
![3 \cdot 8 = 3 (8)](https://tex.z-dn.net/?f=3%20%5Ccdot%208%20%3D%203%20%288%29)
then;
our equation becomes:
![5(8) \div 5+ 3(8)](https://tex.z-dn.net/?f=5%288%29%20%5Cdiv%205%2B%203%288%29)
therefore, the expression which replace the box to form an equation is ![5(8) \div 5+ 3(8)](https://tex.z-dn.net/?f=5%288%29%20%5Cdiv%205%2B%203%288%29)