Answer:
x = 1/4
Step-by-step explanation:
To solve, we need to find the constant of proportionality (k) which is defined by the equation:
yx = k
Therefore when y= 16,
You want to find
P(1000 < X < 3000)
where X is normally distributed with mean 1751 and standard deviation 421. Transform X to Z, so that it follows the standard normal distribution with mean 0 and standard deviation 1 using the relation
X = 1751 + 421Z ==> Z = (X - 1751)/421
Then
P(1000 < X < 3000) = P((1000 - 1751)/421 < (X - 1751)/421 < (3000 - 1751)/421)
… ≈ P(-1.783 < Z < 2.967)
… ≈ P(Z < 2.967) - P(Z < -1.783)
… ≈ 0.9985 - 0.0373
… ≈ 0.9612
so that approximately 96.1% of the students fall in this income range.
-3+9= 6 hope this helped!
Answer:
2/7
then 2/14
Step-by-step explanation:
Let P(H)=p be the probability of one head. In many scenarios, this probability is assumed to be p=12 for an unbiased coin. In this instance, P(H)=3P(T) so that p=3(1−p)⟹4p=3 or p=34.
You are interested in the event that out of three coin tosses, at least 2 of them are Heads, or equivalently, at most one of them is tails. So you are interested in finding the likelihood of zero tails, or one tails.
The probability of zero tails would be the case where you only received heads. Since each coin toss is independent, you can multiply these three tosses together: P(H)P(H)P(H)=p3 or in your case, (34)3=2764.
Now we must consider the case where one of your coin flips is a tails. Since you have three flips, you have three independent opportunities for tails. The likelihood of two heads and one tails is 3(p2)(1−p). The reason for the 3 coefficient is the fact that there are three possible events which include two heads and one tails: HHT,HTH,THH. In your case (where the coin is 3 times more likely to have heads): 3(34)2(14)=2764.
Adding those events together you get p3+3(p2)(1−p)=5464. Note that the 3 coefficient
