When using a two sample test of means with equal standard deviations, then T distribution is the best option for it.
Sample test:
Sample test refers a method of reaching a conclusion to reject or support the claims based on sample data.
Given,
Here we need to identify in which of the following are assumptions made when using a two sample test of means with equal standard deviations.
While we looking into the options for this one, then we have identified that, the sampling distribution of the difference of means is a T distribution.
So, the populations have equal but unknown standard deviation's.
Therefore, we "pool" the sample standard deviation.
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The answer is 2/5 and 3/5 as they both add up to 1.00 or 100%
Answer:
25%
Step-by-step explanation:
From the question given above, the following data were obtained:
Total shaded checker board (T) = 32
Shaded checker board with pieces (S) = 8
Percentage of Shaded checker board with pieces =?
We can obtain the percentage of checker board with pieces as follow:
Percentage of Shaded checker board with pieces = Shaded checker board with pieces / Total shaded checker board × 100
= S / T × 100
= 8 / 32 × 100
= 25%
Therefore, percentage of Shaded checker board with pieces is 25%
Please, use ( ) around fractional coefficients:
<span>1/3x-9y, 6/5x-15y should (probably) be written as (1/3) - 9y and (6/5)x - 15 y.
The LCD here is 3(5) = 15.
Mult. </span><span>1/3x-9y and 6/5x-15y by 15, and then divide the result by 15:
</span>15 [ (1/3)x - 9y and (6/5)x - 15 y ]
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15
This simplifies to:
5x - 135y and 18x - 225y
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15 15
These fractions are equivalent to the original ones, but now the common denominator, 15, is evident.
Answer:D
Step-by-step explanation:
