Hi, the answer to this would be x6/9. I'm assuming the x4/3 and x2/3 are fractions and the x's aren't exponents. Now how I got x6/9 is shown here.
1st Step: Started off by regrouping the terms
1/3x3 x^4x^2
2nd Step: we can easily simplify 3x3 to just 9. And now we're left with 1/9x^4x^2
3rd Step: Now we can simplify the 1/9 to just x^4x^2/9
4th Step: Now we can use the product rule which is simple. So We add the exponents and simplify it to just one exponent. So x4+2=6 that simplifies to just x^6.
Final Answer: x^6/9.
Hope this helped you :)
Answer:
Grouping and then use the common factor. Therefor the answer is 6(2p+5)
Step-by-step explanation:
I hope this helps.
Given that B is the midpoint of line AC and line BC is congruent to line DE.
The following statements and reasons, proves that line AB is congruent to line DE.
Statement Reasons
1. B is the midpoint of line AC Given
2. Line AB is congruent to line BC. Midpoint of a line segment
3. Line BC is congruent to line DE Given
4. Line AB is congruent to line DE Transitive property
(a)36 = 6
(b)81 = 9
(c)256 = 16
(d)196 = 14
Answer:
7.3% of the bearings produced will not be acceptable
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:

Target value of .500 in. A bearing is acceptable if its diameter is within .004 in. of this target value.
So bearing larger than 0.504 in or smaller than 0.496 in are not acceptable.
Larger than 0.504
1 subtracted by the pvalue of Z when X = 0.504.



has a pvalue of 0.9938
1 - 0.9938= 0.0062
Smaller than 0.496
pvalue of Z when X = -1.5



has a pvalue of 0.0668
0.0668 + 0.0062 = 0.073
7.3% of the bearings produced will not be acceptable