What is the degree for the polynomial below ? 2x^(2)+3x + 1

Samples of emissions from three suppliers are classified for conformance to air-quality specifications. The results from 100 sam

tigry1 [53]

Answer:

$P(A) = \frac{30}{100}$

$P(B) = \frac{77}{100}$

$P(A\ n\ B) = \frac{22}{100}$

$P(A\ u\ B) = \frac{85}{100}$

Step-by-step explanation:

Given

See attachment for proper format of table

$n = 100$ --- Sample

A = Supplier 1

B = Conforms to specification

Solving (a): P(A)

Here, we only consider data in sample 1 row.

In this row:

$Yes = 22$ and $No = 8$

So, we have:

$n(A) = Yes + No$

$n(A) = 22 + 8$

$n(A) = 30$

P(A) is then calculated as:

$P(A) = \frac{n(A)}{Sample}$

$P(A) = \frac{30}{100}$

Solving (b): P(B)

Here, we only consider data in the Yes column.

In this column:

$(1) = 22$ $(2) = 25$ and $(3) = 30$

So, we have:

$n(B) = (1) + (2) + (3)$

$n(B) = 22 + 25 + 30$

$n(B) = 77$

P(B) is then calculated as:

$P(B) = \frac{n(B)}{Sample}$

$P(B) = \frac{77}{100}$

Solving (c): P(A n B)

Here, we only consider the similar cell in the yes column and sample 1 row.

This cell is: [Supplier 1][Yes]

And it is represented with; n(A n B)

So, we have:

$n(A\ n\ B) = 22$

The probability is then calculated as:

$P(A\ n\ B) = \frac{n(A\ n\ B)}{Sample}$

$P(A\ n\ B) = \frac{22}{100}$

Solving (d): P(A u B)

This is calculated as:

$P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)$

This gives:

$P(A\ u\ B) = \frac{30}{100} + \frac{77}{100} - \frac{22}{100}$

Take LCM

$P(A\ u\ B) = \frac{30+77-22}{100}$

$P(A\ u\ B) = \frac{85}{100}$

8 0

3 years ago

I have no idea what the answers are

lara [203]

Hello from MrBillDoesMath!

Answer:

Top line: y = (2/3)x + 2

Bottom line: y = (2/3)x -1

Discussion:

The graph provided is hard to read but I did the best I could.

The top line appears to pass through the points (0,2) and (-3,0)

For this line

m = change y /change x = (0-2)/(-3-0) = -2/-3 = +2/3. So

y = mx + b => y = (2/3) x+ b. As the line passes through (0,2) set x = 0, y= 2 in y = (2/3)x + b =>

2 = (2/3) 0 + b => b = 2

Therefore y = (2/3)x + 2

The bottom line appears to pass through the points (0,-1) and (3,1)

For this line

m = change y /change x = (1-(-1)) /(3-0) = +2/-3. So

y = mx + b => y = (2/3) x+ b. As the line passes through (0,-1) set x = 0, y= -1 in y = (2/3)x + b =>

-1 = (2/3) 0 + b => b = -1

Therefore y = (2/3)x + -1

Thank you,

MrB

5 0

3 years ago

Find all solutions in the integers of the equation ||x−3|−5| = 10, where the bars indicate absolute value.

Helga [31]

Answer:

x = 18 or x = -12.

Step-by-step explanation:

||x-3|-5| = 10 only if |x-3|-5 = 10 or |x-3|-5 = -10, i.e., if |x-3|=15 or |x-3|=-5; but |x-3| cannot be equal to -5, because |x-3| should be a non-negative value. Therefore, the first equation is true only if |x-3|=15. |x-3| = 15 only if x-3 = 15 or x-3 = -15, i.e., x = 18 or x = -12. We can verify this in the following way: ||18-3|-5|=||15|-5|=|10|=10 and ||-12-3|-5|=||-15|-5|=|15-5|=|10|=10. This verify that our solution is correct.

3 0

4 years ago

Y=20+6x x=150-4x Solve with equal values method

katen-ka-za [31]

Step-by-step explanation:

Identify the method that will be used to solve for x for each equation.

4x=20 division property fo equality

x-11=9 addition property of eqaulity

5x+6x=22 Combining Like terms

5(x-2)=30 distibute property

HOPE THIS HELPS!!!!!!!!

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4 0

3 years ago

A pencil and a ruler cost $1.50 together. The ruler costs $1.00 more than the pencil. How much does the pencil cost?

Vlad1618 [11]

So we want to know how much does the pencil costs if the ruler costs 1$ more than the pencil and together they cost 1.5$. So lets start with converting dollars into cents. Lets turn this into an equation. Ruler is x and the pencil is y. So together ruler and pencil cost 150cents or: x+y=150 cents. The ruler costs 100 cents more than the pencil, or: (x+100 cents ) + y = 150 cents. So we see that we need to put the values of x=25 cents and y=25 cents to get the ruler to cost 125 cents which is 100 cents or 1$ more than the pencil. So the pencil costs 25 cents.

7 0

3 years ago