<span>No.
To check this yourself, you need the denominators to be the same to be able to easily compare the two.
For example, does 5/8 = 40/64?
1. Determine what you would need to do to the denominator in 5/8 to make it 64. (Multiply it by 8)
2. Find what fraction is equal to 1 with a denominator of 8. (8/8)
3. Multiply the fraction 5/8 by the one you just found (8/8), numerator times numerator, denominator times denominator.
4. Compare the answer with the second fraction.
It is important that when you multiply the denominator by any number you multiply the numerator by the same number. This is to preserve the fraction's value. This works because any number divided by itself is equal to 1, AND when you multiply any number by 1 (whether 1 is in the form of 1 or 4/4 or 8/8 or 234/234), the answer is always equal to the original number.
Another way to check would be to simply enter 1/2 into a calculator, write down the answer. Next enter 5/8 into a calculator. If the answers are the same, they are equal.</span><span>
</span>
The answer is 6480.
Length: 6x3=18
Width: 4x3=12
Height:10x3=30
LengthxWidthxHeight=18x12x30=6480
Answer: it increased by $403. He ended with $6603
Step-by-step explanation: since he started off with $6200, you would move the decimal over two places (.065) and then multiply 6200 and .065 to get $403.
Answer:
- Two sided t-test ( d )
- 0.245782 ( c )
- Since P-value is too large we cannot conclude that the students’ weight are different for these two schools. ( c )
- The test is inconclusive; thus we cannot claim that the average weights are different. ( b )
Step-by-step explanation:
1) Test performed is a Two sided test and this because we are trying to determine the mean difference between two groups irrespective of their direction
<u>2) Determine the P-value ( we will use a data-data analysis approach on excel data sheet while assuming Unequal variances )</u>
yes No
Mean 94.47059 89.76471
Variance 173.2647 95.19118
Observations 17 17
df 30
t Stat 1.184211
P(T<=t) one-tail 0.122814
t Critical one-tail 1.697261
P(T<=t) two-tail 0.245782
Hence The p-value = 0.245782
3) Since P-value is too large we cannot conclude that the students’ weight are different for these two schools.
4) The test is inconclusive; thus we cannot claim that the average weights are different.
