Spencer has 1/3 of nuts. He divides the nuts equally into 4 bags. What fraction of a pound of nuts is it each bag?
1 answer:
You might be interested in
Using the <em>normal distribution and the central limit theorem</em>, it is found that the probability the mean cost of the weddings is more than the mean cost of the showers is of 0.9665.
<h3>Normal Probability Distribution</h3>In a normal distribution with mean and standard deviation
, the z-score of a measure X is given by:
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation
.
- When two variables are subtracted, the mean is the subtraction of the means, while the standard error is the square root of the sum of the variances.
For each sample, they are given by:
For the distribution of differences, we have that:
The probability the mean cost of the weddings is more than the mean cost of the showers is P(X > 0), that is, <u>one subtracted by the p-value of Z when X = 0</u>, hence:
By the Central Limit Theorem
has a p-value of 0.0335.
1 - 0.0335 = 0.9665.
More can be learned about the <em>normal distribution and the central limit theorem</em> at brainly.com/question/24663213
Answer:
The engine should be turned off at t = 1
Step-by-step explanation:
The detailed and step by step calculation is shown in the attachment below
Hello,
Vertices are on a line parallele at ox (y=-3)
The hyperbola is horizontal.
Equation is (x-h)²/a²- (y-k)²/b²=1
Center =middle of the vertices=((-2+6)/2,-3)=(2,-3)
(h+a,k) = (6,-3)
(h-a,k)=(-2,-3)
==>k=-3 and 2h=4 ==>h=2
==>a=6-h=6-2=4 (semi-transverse axis)
Foci: (h+c,k) ,(h-c,k)
h=2 ==>c=8-2=6
c²=a²+b²==>b²=36-4²=20
Equation is:
Vertices are on a line parallele at ox (y=-3)
The hyperbola is horizontal.
Equation is (x-h)²/a²- (y-k)²/b²=1
Center =middle of the vertices=((-2+6)/2,-3)=(2,-3)
(h+a,k) = (6,-3)
(h-a,k)=(-2,-3)
==>k=-3 and 2h=4 ==>h=2
==>a=6-h=6-2=4 (semi-transverse axis)
Foci: (h+c,k) ,(h-c,k)
h=2 ==>c=8-2=6
c²=a²+b²==>b²=36-4²=20
Equation is: