ben is filling up his cylinder shaped pool up to 80% of its capacity . If his is 6 feet deep and has a diameter of 18 feet, how
1 answer:
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<u>Answer:</u>
The value of f(x) = when x=3 is 19, hence the correct option is ‘d’.
<u>Solution:</u>
A two degree polynomial equation is given in the question.
The equation is:
We need <em>to find the value of f(x) when the value of x is 3. </em>
To solve a question like this we have to substitute the given value of x in the given equation and then simplify it.
Now let's substitute the value of x in the given equation.
First we square the number 3 and then multiply it with 2 and then add 1.
This is done because of the BODMAS rule, in which multiplication is given higher preference as compared to addition.
On solving the equation we get,
F(x)=2×9+1
F(x)=18+1
F(x)=19.
Therefore the value of f(x) when x=3 is 19.
Hence the correct option is ‘d’.
Answer:
There is a 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard deviation of 2.1-cm. This means that .
For shipment, 9 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.
By the Central Limit Theorem, since we are using the mean of the sample, we have to use the standard deviation of the sample in the Z formula. That is:
This probability is 1 subtracted by the pvalue of Z when .
has a pvalue of 0.9992. This means that there is a 1-0.9992 = 0.0008 = 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.