Write the expression as a polynomial: (5–2a+a^2)(4a^2–3a–1)
1 answer:
Answer: 4a^4 - 11a^3 + 25a^2 - 13a - 5
Explanation:
Check out the attached image below for the box method diagram. It's basically a 3 by 3 box showing all the 9 terms, some of which combine together (only the like terms though).
Each inner box or inner cell is the result of multiplying the outer terms. For example, in row 2, column 2 we have the outer terms -3a and -2a multiply to get 6a^2. Figure 1 shows the full table filled out while figure 2 has that same table highlighted with different colors to show the like terms.
The like terms are:
-15a and 2a, which combine to -13a
20a^2, 6a^2 and -a^2 which combine to 25a^2
-8a^3 and -3a^3 which combine to -11a^3
Those results are added to the other terms but we do not combine them (as they are not like terms). That's how we end up with the final answer. This is a quartic polynomial as the degree (largest exponent) is 4. There are five terms in this polynomial when fully simplified.
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Answer:
a) This is the p-value of Z when X = A subtracted by the p-value of Z when X = B.
b) P-value of Z when X = 76 subtracted by the p-value of X = 68.
c) Because the underlying distribution(pulse rates of females) is normal.
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.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question:
Mean , standard deviation
.
standard deviation of beats per minute.
a. If 1 adult female is randomly selected, find the probability that her pulse rate is between A beats per minute and B beats per minute.
This is the p-value of Z when X = A subtracted by the p-value of Z when X = B.
b. If 4 adult females are selected, find the probability that they have pulse rates with a mean between 68 beats per minute and 76 beats per minute.
Sample of 4 means that we have
The formula for the z-score is:
This probability is the p-value of Z when X = 76 subtracted by the p-value of X = 68.
c. Why can the normal distribution be used in heartbeat even the sample side does not exceed 30?
Because the underlying distribution(pulse rates of females) is normal.
9514 1404 393
Answer:
x = -7, x = 9
Step-by-step explanation:
We presume your equation is ...
x² -2x -63 = 0
Factors of -63 that have a sum of -2 are -9 and +7. Then the factored equation is ...
(x -9)(x +7) = 0
Solutions make the factors zero.
x -9 = 0 ⇒ x = 9
x +7 = 0 ⇒ x = -7
The solutions to the quadratic equation are x = -7 and x = 9.
