3 boxes of computer diskettes at $8.99 per box = (3 * 8.99 = 26.97 )
2 packages of pens at $ 2.50 per package = (2 * 2.50 = 5.00 )
6 boxes of paper at $ 22.95 per box = (6 * 22.95 = 137.70)
1 printer ribbon at $25 each. = (1 * 25 = 25)
Add up (26.97+5.00+137.70+25 ) and the answer is = 194.67
Answer:
66%
Step-by-step explanation:
Given :
Age Percentile
75 97
65 90
55 78
45 58
35 37
25 13
20 66
To Find :The percentage of drivers who are younger than 20 is _____%.
Solution:
Refer the given table
The percentile corresponding to age 20 is 66.
Percentile : A data item is said to be in nth percentile of distribution if n% of the distribution are less than that particular data item.
So, drivers who are younger than 20 are said to be in 66 percentile if 66%of the distribution are less than that particular data item.
Hence The percentage of drivers who are younger than 20 is 66%.
56 miles, The car traveled m miles during the first hour and m squared miles during the second hour. The ratio m/m squared = 1/7 and cross multiplying yields m squared = 7m, so m=7. This means that the total distance traveled was 7+7squared = 56 miles
Answer:
80.0456<
<81.1210
Step-by-step explanation:
-Given the mean, 
and 
, the confidence interval can be calculated using the formula:
#We substitute our values in the formula to solve for CI:
![=\bar x\pm z\times \frac{\sigma}{\sqrt{n}}\\\\=\bar y\pm z_{0.05}\times \frac{s}{\sqrt{72}}\\\\=80.5833\pm 1.645\times \frac{2.77369}{\sqrt{72}}\\\\=80.5833\pm0.5377\\\\=[80.0456,81.1210]](https://tex.z-dn.net/?f=%3D%5Cbar%20x%5Cpm%20z%5Ctimes%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%5C%5C%5C%5C%3D%5Cbar%20y%5Cpm%20z_%7B0.05%7D%5Ctimes%20%5Cfrac%7Bs%7D%7B%5Csqrt%7B72%7D%7D%5C%5C%5C%5C%3D80.5833%5Cpm%201.645%5Ctimes%20%5Cfrac%7B2.77369%7D%7B%5Csqrt%7B72%7D%7D%5C%5C%5C%5C%3D80.5833%5Cpm0.5377%5C%5C%5C%5C%3D%5B80.0456%2C81.1210%5D)
Hence, the confidence interval lies between 80.0456 and 81.1210
