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2 years ago

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

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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

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