12 3/4 miles
Eric is increasing his distance by 2 3/8 miles per day. In 2 more days, his distance will be 2×(2 3/8) = 4 3/4 miles more than on Wednesday.
... 8 + 4 3/4 = 12 3/4 . . . . miles on Friday
_____
Rate of increase
The difference between Tuesday and Monday is ...
... 5 5/8 - 3 2/8 = 2 3/8
The difference between Wednesday and Tuesday is ...
... 8 - 5 5/8 = 2 3/8
Thus, it appears that Eric is jogging 2 3/8 mile more each day than the day before.
Answer:
Step-by-step explanation:
Let's call x the length of each piece in yards
If the ribbon is cut into 8 equal pieces then the sum of the pieces is:
Then:
Now we solve the equation for x
Divide both sides of equality by 8
So the length of each piece is of a yard.
This is: 1.125 yards
Answer: Option 'c' is correct.
Step-by-step explanation:
Since we have given that
We need to find the number of units in order to minimize cost.
We first derivative w.r.t. x,
For critical points:
Now, we will check whether it is minimum or not.
We will find second derivative .
So, it will yield minimum cost.
Minimum cost would be
Hence, At 150 units, minimum cost = $4000
Therefore, Option 'c' is correct.
Could you explain the process to solve this type of problem? I have ten to do and this is only #1. I need to be able to solve it on my own before I continue with my lesson.
Answer:
There is sufficient evidence to warrant rejection of the claim that professional basketball players are born in different months with the same frequency.
Step-by-step explanation:
In this case we need to test whether there is sufficient evidence to warrant rejection of the claim that professional basketball players are born in different months with the same frequency.
A Chi-square test for goodness of fit will be used in this case.
The hypothesis can be defined as:
H₀: The observed frequencies are same as the expected frequencies.
Hₐ: The observed frequencies are not same as the expected frequencies.
The test statistic is given as follows:
The values are computed in the table.
The test statistic value is .
The degrees of freedom of the test is:
n - 1 = 12 - 1 = 11
Compute the p-value of the test as follows:
p-value < 0.00001
*Use a Chi-square table.
p-value < 0.00001 < α = 0.05.
So, the null hypothesis will be rejected at any significance level.
Thus, there is sufficient evidence to warrant rejection of the claim that professional basketball players are born in different months with the same frequency.