Let m∠CLN = x. Then m∠ALM = 3x, and m∠A = 90°-x, m∠C = 90°-3x.
The sum of angles of ∆ABC is 180°, so we have
... 180° = 40° + m∠A + m∠C
Using the above expressions for m∠A and m∠C, we can write ...
... 180° = 40° + (90° -x) + (90° -3x)
... 4x = 40° . . . . . . . . . add 4x-180°
... x = 10°
From which we conclude ...
... m∠C = 90°-3x = 90° - 3·10° = 60°
The ratio of CN to CL is
... CN/CL = cos(∠C) = cos(60°)
... CN/CL = 1/2
so ...
... CN = (1/2)CL
Answer:
(1) Correct option (A).
(2) The probability that Aadi will get Tails is 0.40.
Step-by-step explanation:
The information provided is:
The probability of tail in both cases is:
Here,
P (T|E) implies the probability of tail in case of Eric's experiment.
P (T|S) implies the probability of tail in case of Sue's experiment.
(1)
Now, it is given that Aadi is going to throw the coin once.
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
In this case we need to compute the probability of Aadi getting Tails in a single toss.
As Sue uses a larger number of trials in the experiment, i.e. n = 50 > 30 times, according to the Central limit theorem, Sue's estimate is best because she throws it .
Thus, the correct option is (A).
(2)
As explained in the first part that Sue's estimate is best for getting a tail, the probability that Aadi will get Tails when he tosses the coin once is:
Thus, the probability that Aadi will get Tails is 0.40.
Answer:
a)
b)
c)
d)
e)
Step-by-step explanation:
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Let X the random variabl length of advertisements produced by Majesty Video Production Inc. We know from the problem that the distribution for the random variable X is given by:
We take a sample of n=16 . That represent the sample size.
a. What can we say about the shape of the distribution of the sample mean time?
From the central limit theorem we know that the distribution for the sample mean is also normal and is given by:
b. What is the standard error of the mean time?
The standard error is given by this formula:
c. What percent of the sample means will be greater than 31.25 seconds?
In order to answer this question we can use the z score in order to find the probabilities, the formula given by:
And we want to find this probability:
d. What percent of the sample means will be greater than 28.25 seconds?
In order to answer this question we can use the z score in order to find the probabilities, the formula is given by:
And we want to find this probability:
e. What percent of the sample means will be greater than 28.25 but less than 31.25 seconds?"
We want this probability: