Answer:
For this case if we want to conclude that the sample does not come from a normally distributed population we need to satisfy the condition that the sample size would be large enough in order to use the central limit theoream and approximate the sample mean with the following distribution:

For this case the condition required in order to consider a sample size large is that n>30, then the best solution would be:
n>= 30
Step-by-step explanation:
For this case if we want to conclude that the sample does not come from a normally distributed population we need to satisfy the condition that the sample size would be large enough in order to use the central limit theoream and approximate the sample mean with the following distribution:

For this case the condition required in order to consider a sample size large is that n>30, then the best solution would be:
n>= 30
Answer:
The correct answer is - 625.36.
Step-by-step explanation:
Given:
Fixed mothly plan - 616
free minutes - 150
Used minutes - 170
Free text - 150
Used text - 182
Charging amount over free limit - 18p
Solution:
Number of minutes over the limit - 170 - 150 = 20
Number of text over the limit - 182 - 150 = 32
So the extra amount that will be add to the monthly charge would be -
(20*0.18) +(32*0.18)
and the total charge of the month would be -
= 616+(20*0.18) +(32*0.18)
= 616+3.60+5.76
= 625.36
Thus, the correct answer would be - 625.36
Answer: it’s 20m long
Step-by-step explanation:
I can’t explain .—.
ratio of the diagonal to the side of the rectangle is 17/8 or 17:8
Step-by-step explanation:
Length of Diagonal = 34 inches
Length of side = 16 inches
Base= 30 inches
We need to find the ratio of the diagonal to the side of the rectangle.
ratio of the diagonal to the side of the rectangle = 
Putting values:

So, ratio of the diagonal to the side of the rectangle is 17/8 or 17:8
Keywords: ratio
Learn more about ratio at:
#learnwithBrainly
Answer:

Step-by-step explanation:
Given the equation;

Rearranging the equation, we have;

Lowest common multiple (LCM) of S and T is ST.

Cross-multiplying, we have;

Making R, the subject of formula;
