Answer:
0.3085 = 30.85% probability that a randomly selected pill contains at least 500 mg of minerals
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean 
and standard deviation 
, the z-score of a measure X is given by:
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean 490 mg and variance of 400.
This means that 
What is the probability that a randomly selected pill contains at least 500 mg of minerals?
This is 1 subtracted by the p-value of Z when X = 500. So
has a p-value of 0.6915.
1 - 0.6915 = 0.3085
0.3085 = 30.85% probability that a randomly selected pill contains at least 500 mg of minerals
Answer:
x=7.5
y=1.5
Step-by-step explanation:
subtract x from both sides
combine like terms
subtract 1.8 from both sides
divide both sides by 0.3
x=7.5
y=1.5
<h2>
Answer with explanation:</h2>
We are given a semi-ellipse gate whose dimensions are as follows:
Height of 20 feet and a width of 15 feet.
Now, if a truck is loaded then:
Height of truck is: 12 feet and a width of truck is: 16 feet
The truck won't pass through the gate since the width of truck is more than that of the gate.
When the truck is not loaded then:
Height of truck is: 12 feet and a width of truck is: 10 feet
The truck would easily pass through the gate since, the dimensions of truck are less than that of the gate.
Answer:
Step-by-step explanation:
Given
Required
Determine the value of T
Multiply both sides by 2
Divide through by 6
Answer:
Cov(X, Y) =0.029.
Step-by-step explanation:
Given that :
The noise in a particular voltage signal has a constant mean of 0.9 V. that is μ = 0.9V ............(1)
Also, the two noise instances sampled τ seconds apart have a bivariate normal distribution with covariance.
0.04e–jτj/10 ............(2)
Having X and Y denoting the noise at times 3 s and 8 s, respectively, the difference of time = 8-3 = 5seconds.
That is, they are 5 seconds apart,
τ = 5 seconds..............(3)
Thus,
Cov(X, Y), for τ = 5seconds = 0.04e-5/10
= 0.04e-0.5 = 0.04/√e
= 0.04/1.6487
= 0.0292
Thus, Cov(X, Y) =0.029.
