Answer:
28 miles
Step-by-step explanation:
Let w stand for the number of miles Derrick runs each week. Then after 4 weeks of training, he wants the total mileage to be 100 miles.
16 + 3w = 100
3w = 84 . . . . . . . subtract 16
w = 28 . . . . . . . . divide by 3
Derrick will run 28 miles each week to achieve his training goal.
Answer:
-1/60
Step-by-step explanation:

= 
=
(find LCM of demoninators)
= 
= 
= -1/60
Answer:
That looks hard but idk how to answer it
Step-by-step explanation:
We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:

The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:
![CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack%20x-Z_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%2Cx%2BZ_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%5Crbrack)
Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:
![CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack30.0-Z_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%2C30.0%2BZ_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%5Crbrack)
Where (from tables):

Finally, the interval at 98% confidence level is:
Answer:

Step-by-step explanation:
Given,
{ 1 , 4 } , { 3 , 5 } , { - 1 , 2 }
To find : Range
<u>Range are the set of all y - co-ordinates .</u>
So,
Answer : { 2 , 4 , 5 }
Hope I helped!
Best regards!