Yes if you would like them to be considered that way depending on you x axis degree
Answer:
c) 2
Step-by-step explanation:
a prime number is a number which is only completely divisible by 1 and itself
Answer:
length = 31 ft
width = 19 ft
Explanation:
Assume that the width of the rectangle is w.
We are given that the length is 12 ft more than the width. This means that the length of the rectangle is w + 12
The perimeter of the rectangle in this case would be:
perimeter = 2 (length + width)
perimeter = 2 (12 + w + w)
perimeter = 24 + 4w
Assume that Dan would use all 100 ft of fencing to surround the yard. This would mean that the largest perimeter is 100 ft.
Therefore:
perimeter = 24 + 4w
100 = 24 + 4w
4w = 100 - 24
4w = 76
w = 19 ft
Since we have calculated that the width of the yard is 19 ft, we can substitute to get the length as follows:
length = 12 + w
length = 12 + 19
length = 31 ft
Hope this helps :)
Answer:
55.74
Step-by-step explanation:
<em>Since the culture was tested and then the countdown started, you need to multiply the initial value by 3.5. Since it is tested once every 15 minutes, it will be tested two more times by the time 30 minutes is up; a total of 3 times tested.</em>
1.3 x 3.5 = 4.55
4.5 x 3.5 = 15.925
15.925 x 3.5 = 55.7375
Answer:
The approximated value of the standard deviation is 0.35.
Step-by-step explanation:
According to the Central Limit Theorem if we have an unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.
Then, the mean of the distribution of sample means is given by,

And the standard deviation of the distribution of sample means is given by,

The information provided is:
<em>n</em> = 100
<em>σ</em> = 3.5
<em>μ</em> = 66
As the sample size is quite large, i.e. <em>n</em> = 100 > 30, the central limit theorem can be applied to approximate the sampling distribution of sample mean by the Normal distribution.
Then the approximated value of the standard deviation of sampling distribution of sample mean is:


Thus, the approximated value of the standard deviation is 0.35.