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Anastasy [175]
2 years ago
6

Find the point of intersection of the lines:

Mathematics
2 answers:
avanturin [10]2 years ago
8 0

Answer:

the answer is a

Step-by-step explanation:

Marta_Voda [28]2 years ago
3 0
The answer is A my guy hope this helps
You might be interested in
2x^3 + 3x^2 -11x -6 divided by x - 2
Gnesinka [82]

p(x)= x-2

g(x)= 2x^3 + 3x^2 - 11x - 6

first we have to find the zero of the polynomial of x-2

p(x)= x-2 = 0

         x=2

therefore,

p(x)= 2x^3 + 3x^2 - 11x - 6

p(2)= 2*2^3 + 3*2^2 - 11*2 - 6

      = 2*8 + 3*4 - 11*2 - 6

      = 16 + 12 - 22 - 6

      = 28-28

     = 0

Hope it helped u, ^_^.

3 0
3 years ago
Please<br>Help me I don't get this! ​
Vikentia [17]
The 4 in and the 6 in are the same but the difference between the two triangles are 1.5

6:4 = 1.5
So if 12 = 1.5 (?)
The (?) = 8
4 0
2 years ago
(c) If a pizza contains 8 slices and there are 4
slavikrds [6]
2 slices per person
7 0
3 years ago
Read 2 more answers
Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be
hodyreva [135]

Answer:

\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305

s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108

\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105

s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221

The absolute difference is:

Abs = |340.221-115.108|= 225.113

If we find the % of change respect the before case we have this:

\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305

And the sample deviation with the following formula:

s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105

And the sample deviation with the following formula:

s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

Abs = |340.221-115.108|= 225.113

If we find the % of change respect the before case we have this:

\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%

So then is a big change.

8 0
3 years ago
Can someone help me with this question?
belka [17]
Option 3 (1/4 x 60) since 59 is close to 60 and 25% can be rewritten as 1/4
4 0
3 years ago
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