Select the correct responses in the table. The relationship between two numbers is described below, where x represents the first
number and y represents the second number. The square of the first number is equal to the sum of the second number and 16. The difference of 4 times the second number and 1 is equal to the first number multiplied by 7. Select the equations that form the system that models this situation. Then, select the solution(s) of the system.
1 answer:
9514 1404 393
Answer:
- x² = y + 16
- 4y - 1 = 7x
- (5, 9)
Step-by-step explanation:
Writing equations from words is mostly a matter of understanding what the English words mean.
If the first number is x, then the square of the first number is x². That is said to be equal to the sum of the second number (y) and 16:
x² = y + 16 . . . . the first equation
The wording "the difference of A and B" is usually intended to be interpreted to mean A-B. Here, we have the difference of 4 times the second number (4y) and 1, so ...
4y -1
This is equal to the first number (x) multiplied by 7, so ...
4y -1 = 7x . . . . the second equation
__
These can be solved a variety of ways. When no method is specified, I like to use a graphing calculator. It shows the only integer solution for this pair of equations is ...
(x, y) = (5, 9)
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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.