Simplify each expression below as much as possible. 6x + 2 - 1 - 4x - 3 - 2x + 2

Answers

Answer 1
Answer: collect like terms
0+2-1-3+2
calculate the sum or difference
0+0
remove 0

solution: 0


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Eric jog 3 1/4 mile on Monday, 5 5/8 miles on Tuesday, and 8 miles on Wednesday. Suppose he continues the pattern for the remainder of the week. How far will Ereck jog on Friday

Answers

Answer:

12 3/4 miles

Step-by-step explanation:

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.

Are they equal or diffrent angles

Answers

equal because the parallel lines :)

Nine yards of ribbon are cut into 8 equal pieces what is the length of each piece of ribbon. Write a division expression to represent the problem and solve

Answers

Answer:

x=(9)/(8)\ yards

x=1.125\ yards

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:

x+x+x+x+x+x+x+x=8x

Then:

8x=9\ yards

Now we solve the equation for x

8x=9\ yards

Divide both sides of equality by 8

(8)/(8)x=(9)/(8)\ yards

x=(9)/(8)\ yards

So the length of each piece is (9)/(8) of a yard.

This is: 1.125 yards

A company’s total cost from manufacturing and selling x units of their product is given by: y = 2x2 – 600x + 49,000. How many units should be manufactured in order to minimize cost, and what is the minimum cost?a. (100, $9,000)b. (125, $3,250)c. (150, $4,000)d. (170, $4,800)e. (200, $9,000)

Answers

Answer: Option 'c' is correct.

Step-by-step explanation:

Since we have given that

y=2x^2-600x+49000

We need to find the number of units in order to minimize cost.

We first derivative w.r.t. x,

y'=4x-600

For critical points:

y'=0\n\n4x-600=0\n\n4x=600\n\nx=(600)/(4)\n\nx=150

Now, we will check whether it is minimum or not.

We will find second derivative .

y''=4>0

So, it will yield minimum cost.

Minimum cost would be

y(150)=2(150)^2-600* 150+49000\n\ny(150)=\$4000

Hence, At 150 units, minimum cost = $4000

Therefore, Option 'c' is correct.

(Algebra II) Solve the equation or formula for the indicated variable. a = 4b5h, for h

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.

Answers

Divide by A and b5 to both sides. You can rule out option A and B. You're dividing both sides by 4b5 to isolate h by itself. Remember that when you didivde by same term aa=1 you get ONE. so h is alone. The answer is D.

An author argued that more basketball players have birthdates in the months immediately following July​ 31, because that was the age cutoff date for nonschool basketball leagues. Here is a sample of frequency counts of months of birthdates of randomly selected professional basketball players starting with​ January: 390​, 392​, 360​, 318​, 344​, 330​, 322​, 496​, 486​, 486​, 381​, 331 . Using a 0.05 significance​ level, is there sufficient evidence to warrant rejection of the claim that professional basketball players are born in different months with the same​ frequency? Do the sample values appear to support the​ author's claim?

Answers

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:

 \chi^(2)=\sum{((O-E)^(2))/(E)}

The values are computed in the table.

The test statistic value is \chi^(2)=128.12.

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.