( c^2 )^3 = 64 <=>( c^2 )^3 = 4^3 <=> c^2 = 4 <=> c = 2 or c = -2.
So the answer choices are
1. a-b=even
2. a and b are not odd
3. a and b are odd
4. a-b=even
5. a-b=not even
even=not odd
not even=odd so choice 5 is really a-b=odd
basically choice 4 and 1 are the same so we cross one out
so
the problem said that a and b are odd so therefor choice2 is wrong and choice 3 is correct
then both are odd
odd-odd=even because
an even number is represented as 2n where n is an integer
an odd number can be represented as 2n+1 so assume you have 2 odd numbers 2 away from each other so odd and (odd+2)
odd+2-odd=2n+1+2-(2n+1)=2n+1+2-2n-1=2n-2n+1-1+2=2
you are left with odd
using integers
7 and 11
11-7=4
even
so odd-odd=even, it depends on weather you consider 0 odd or even
so the asnwers are:
a and b are odd
a-b is not an even integer
Answer:
14
Step-by-step explanation:
Since all sides are half of greater triangle.Then 7 × 2 = 14 is length of greater hypotenuse.
Suppose two triangles with similar conditions
1^2+1^2=2
Hypotenuse=2^1/2
We assume another triangle which is exactly double of previous triangle I.e
2^2+2^2=8
Hypotenuse=2×2^1/2
Which is exactly twice of previous triangle.
Hence proved hyootenuse is double of smaller triangle I.e 14
Answer:
3.6cm
Step-by-step explanation:
Given data
We are given that the dimension of the photo is
Height= 15cm
Width= 9cm
That is
15cm height will make 9cm width
hence 6cm height will make x widht
cross multiply
9*6= 15*x
54= 15x
x= 54/15
x= 3.6cm
Hence the width will be 3.6cm
Answer:
A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of 
, and a confidence level of 
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of 
.
The margin of error of the interval is given by:
In this problem, we have that:
99.5% confidence level
So 
, z is the value of Z that has a pvalue of 
, so 
.
Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.07?
This is n when M = 0.07. So
A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07
