Simplifying Nonperfect Roots
1 answer:
Short Answer:Choice C [Third one down]
Remark
Start with the number so I can talk about the basic principle at work. The way to work it is to factor 162 into prime factors and hope there are at least 3 that are the same.
162 = 2 * 81
162 = 2 * 3 * 3 * 3 * 3
Now When you take the cube root of that, you get![\sqrt[3]{2*2*2*2*3}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B2%2A2%2A2%2A2%2A3%7D%20)
Here's the rule for a cube root. For every 3 prime factors under the cuberoot sign, you take out one and throw the other two away. So for cube root of 81,
you would factor it as ∛(81) = ∛(3 * 3 * 3 * 3) = 3∛3.
So out of four 3s under the cube root sign, you have 1 outside the root sign and one inside the root sign. 2 of the four 3s have been thrown away.
Continuation.
X first
You want 2 xs outside the root sign
∛(x * x * x * x * x *x ) = (x * x) You have thrown away 2 xs for every x outside the cube root sign
C = 6 There are no left overs.
Y second
For the ys, you need 1 outside and 2 inside the root sign. that's because you need 5 altogether.
∛(y * y * y * y * y) = y ∛y^2
Answer
c = 6; d = 2
Choice C <<<<<< answer
Remark
Start with the number so I can talk about the basic principle at work. The way to work it is to factor 162 into prime factors and hope there are at least 3 that are the same.
162 = 2 * 81
162 = 2 * 3 * 3 * 3 * 3
Now When you take the cube root of that, you get
Here's the rule for a cube root. For every 3 prime factors under the cuberoot sign, you take out one and throw the other two away. So for cube root of 81,
you would factor it as ∛(81) = ∛(3 * 3 * 3 * 3) = 3∛3.
So out of four 3s under the cube root sign, you have 1 outside the root sign and one inside the root sign. 2 of the four 3s have been thrown away.
Continuation.
X first
You want 2 xs outside the root sign
∛(x * x * x * x * x *x ) = (x * x) You have thrown away 2 xs for every x outside the cube root sign
C = 6 There are no left overs.
Y second
For the ys, you need 1 outside and 2 inside the root sign. that's because you need 5 altogether.
∛(y * y * y * y * y) = y ∛y^2
Answer
c = 6; d = 2
Choice C <<<<<< answer
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Answer:
a) 51.60% probability that in a given year there will be less than 21 earthquakes.
b) 49.35% probability that in a given year there will be between 18 and 23 earthquakes.
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) Find the probability that in a given year there will be less than 21 earthquakes.
This is the pvalue of Z when X = 21. So
has a pvalue of 0.5160.
So there is a 51.60% probability that in a given year there will be less than 21 earthquakes.
b) Find the probability that in a given year there will be between 18 and 23 earthquakes.
This is the pvalue of Z when X = 23 subtracted by the pvalue of Z when X = 18. So:
X = 23
has a pvalue of 0.7611
X = 18
has a pvalue of 0.2676
So there is a 0.7611 - 0.2676 = 0.4935 = 49.35% probability that in a given year there will be between 18 and 23 earthquakes.