What are the coordinates of the circumcwnter of this triangle
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Answer:
Probability of at least 50 obese individuals in our sample is 0.92364 .
Step-by-step explanation:
We are given that a random sample of 300 adults is taken from the state of Colorado, where the rate of obesity is 19.8% .
Let X = Number of obese individuals
Firstly, X ~
For approximating binomial distribution into normal distribution, firstly we have to calculate and
.
Mean of Normal distribution, = n * p = 300 * 0.198 = 59.4
Variance of Normal distribution, = n * p * (1-p) = 300 *0.198 *0.802 = 47.64
So, now X ~ N()
The standard normal z score distribution is given by;
Z = ~ N(0,1)
So, probability of at least 50 obese individuals in our sample = P(X >= 50)
P(X >= 50) = P(X > 49.5) {using continuity correction}
P(X > 49.5) = P( >
) = P(Z > -1.43) = P(Z < 1.43) = 0.92364
Therefore, required probability is 0.92364 .
First box = 14 pounds
Second box = 2x = 28 pounds
Third Box = x+2= 16 pounds
Fourth box = x/2 = 7 pounds
<u>Step-by-step explanation:</u>
Here we have , Sixty-five pounds of candy was divided into four different boxes. The second box contained twice the amount of the first box. The third box contained two more pounds than the first box. The last box contained one-fourth the amount in the second box. We need to find How much candy was in each box. Let's find out:
We have a total of 65 pounds of candy ! Let in first box we have x pounds so , second box contained twice the amount of the first box i.e.
⇒
The third box contained two more pounds than the first box i.e.
⇒
The last box contained one-fourth the amount in the second box i.e.
⇒
Therefore , Sum of pounds of candy are :
⇒
⇒
⇒
⇒
⇒
Therefore , Candy in each box is :
First box = 14 pounds
Second box = 2x = 28 pounds
Third Box = x+2= 16 pounds
Fourth box = x/2 = 7 pounds