Can any one pleasssssseeee help me with algebra home work? I need help on A, B, and C
Can any one pleasssssseeee help me with algebra home work? - 1

Answers

Answer 1
Answer: By looking at this scatter plot you can see that 2 people bowled above their average, 1 person bowled the same number for their average and their highest score (this person bowled 200 as their highest score as well as their average), and the biggest difference between someones high score and their average is one person has 200 for their bowling average, but only 175 for their high score. Everyone else's high scores and averages are about the same.

Related Questions

Is 1-6 correct please help I will give thanks or brainliest
Al makes a bank deposit every third day. He makes a withdrawal every fifth day. He made both a deposit and withdrawal on dec 1. When will he next make a deposit and a withdrawal on the same day?
jess puts 1 2/3 tomatoes in each salad. Dott ate 1 1/2 salads yesterday and 2 7/10 salads today. How many more tomatoes did she eat today? Explain how you solved this problem.
Alexandra purchases two doughnuts and three cookies at a doughnut shop and is charged $3.30. Brianna purchases five doughnuts and two cookies at the same shop for $4.95. All the doughnuts and cookies have the same price. Find the cost of one doughnut and the cost of one cookie.
Tom drank 1 1/4 quarts of water and his sister Jane drank 1.75 quarts of water. Write the amount that Jane drank using afraction. Who drank more water?(Doesn’t have choices)

20% of a number is 64
what is 50% of the number?

explain how you worked this out

Answers

is        %
__  =  __
of      100

64       20%
__   =  __
 x        100

20x= 6,400
Divide.
6,400/20=320.
Half or 50% of 320 is 160.

 

Is 0.602 equvalent to 0.62

                

Answers

No.
0.602 = 0.6020 = 0.60200 etc
0.62 = 0.620 = 0.6200 etc
Adding zeros to the end of a decimal doesn't change the value, the order of the digits does.
No.. 0.62 is a larger number than 0.602

Can you help me if it is equivalent or not eqvalent you can help me with the rest if you want the brainless answer

Answers

10. Not equivalent.

11. Equivalent

12. Not equivalent

13. Not equivalent

14. Equivalent

15. Can't read, but I think not equivalent.

Basically, if they can be multiplied by the same scale factor they are equivalent, else they aren't.

Answer:

10. Not equivalent.

11. Equivalent

12. Not equivalent

13. Not equivalent

14. Equivalent

15. Can't read, but I think not equivalent.

Basically, if they can be multiplied by the same scale factor they are equivalent, else they aren't.

Step-by-step explanation:

what that guy said :0

Use row operations to solve the systemx + y – z = -2
4x – y + z = 7
X – 3y + 2z =-5

Answers

Answer:

  (x, y, z) = (1, 12, 15)

Step-by-step explanation:

As with any set of linear equations, there are many possible routes to a solution. We might simplify the notation a bit by writing the coefficients in an augmented matrix. The columns, left to right, represent the coefficients of x, y, and z, in order, and the constant term.

The row operations we'll use are multiplying a row by a value and adding that result to another row, replacing the other row by the sum.

We can make things a little simpler by writing the second equation first. Then the augmented matrix we're starting with is ...

  \left[\begin{array}{ccc|c}4&-1&1&7\n1&1&-1&-2\n1&-3&2&-5\end{array}\right]

Adding the second row to the first, we get ...

  \left[\begin{array}{ccc|c}5&0&0&5\n1&1&-1&-2\n1&-3&2&-5\end{array}\right]

Dividing the first row by 5 gives ...

  \left[\begin{array}{ccc|c}1&0&0&1\n1&1&-1&-2\n1&-3&2&-5\end{array}\right]

Subtracting this from the second row, and again from the third row, we are left with ...

  \left[\begin{array}{ccc|c}1&0&0&1\n0&1&-1&-3\n0&-3&2&-6\end{array}\right]

Multiplying the second row by 3 and adding that to the third row, we get ...

  \left[\begin{array}{ccc|c}1&0&0&1\n0&1&-1&-3\n0&0&-1&-15\end{array}\right]

Subtracting the third row from the second gives ...

  \left[\begin{array}{ccc|c}1&0&0&1\n0&1&0&12\n0&0&-1&-15\end{array}\right]

Finally, multiplying the last row by -1, we have the solution:

  \left[\begin{array}{ccc|c}1&0&0&1\n0&1&0&12\n0&0&1&15\end{array}\right]

This matrix corresponds to the equations ...

  • x = 1
  • y = 12
  • z = 15

_____

The purpose of our choice of row operations is to make the diagonal elements 1 and the off-diagonal elements 0. That is how we end up with the final equations shown.

As we said, there are many ways to go about this. In general, one can ...

  • if necessary, swap rows until the diagonal term of interest is non-zero. If you are doing this using a computer program, generally you want the diagonal term to have the coefficient with the largest magnitude. When doing this by hand, you may want to arrange the rows to avoid fractions when you do the normalizing.
  • divide the row by the coefficient of the diagonal element to "normalize" the diagonal element to a value of 1
  • zero the other elements in that column by multiplying the row just normalized by the element in another row, then subtracting the product. (The 4th matrix shown above shows this for the first column.)
  • proceed to the next diagonal element and repeat the process until all diagonal elements are 1. If you cannot make all diagonal elements 1, then the system of equations does not have a unique solution. If any row becomes all zeros, the system is "dependent" and has infinite solutions. If a row is zeros except for the rightmost column, the system is "inconsistent" and has no solutions.

What number is 20% of 20

Answers

10% of a number can be found by moving the decimal point over one space to the left.
20 move the decimal point to the left = 2
Now that we know that 10% of 20 is 2, we can find 20% by multiplying 2 by 2.
20% of 20 is 4.
Our answer is 4.

What are the property's for (1.) n times 1=n


(2.) 4 times m = m times 4


(3.) {x + 2} + 5 = x + {2 +5}

Answers

1. Multiplicative Identity Property

2. Commutative Property of Multiplication

3. Commutative Property of Addition

Hope this helped! :D