Ask question

Ask question

All categories

3 years ago

14

In the xy-plane, if (-1, 0) is a solution to the system of inequalities y < x + c and y < -x - d, which of the following m

ust be true about c and d? (10 POINTS + BRAINLIEST ANSWER!!)
A. c = d

B. c = -d

C. c < d

D. d < c

1 answer:

BlackZzzverrR [31]3 years ago

5 0

Answer:

76 87 12 35 67

Step-by-step explanation:vthats right

You might be interested in

Using the technique in the model above, find the missing sides in this 30° 600-90 right triangle.longShort 7Hypotenuse3.571314

dalvyx [7]

The answer to this is 14. To find the hypotenuse from the short do 2S, and for Short to Long it is S Root 3.

Hope this helped! Please Mark as Brainliest!

3 0

4 years ago

the lions won 16 games last year. this year the lions won 20 games. what is the percent increase in number of games the lions wo

jarptica [38.1K]

Last year won the lions 16 games last year and into this year the Lions won 20 games.

as the percentage increase is:

20/16

(20/4) / (16/4)

5/4
<span>
1.25 (125%)</span>

6 0

3 years ago

There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that pl

insens350 [35]

We're told that

$P(A\cap B)=0.15$

$P(A\cup B)^C=0.06\implies P(A\cup B)=0.94$

$P(B\mid A)=P(B^C\mid A)=0.5$

where the last fact is due to the law of total probability:

$P(A)=P(A\cap B)+P(A\cap B^C)$

$\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)$

$\implies 1=P(B\mid A)+P(B^C\mid A)$

so that $B\mid A$ and $B^C\mid A$ are complementary.

By definition of conditional probability, we have

$P(B\mid A)=P(B^C\mid A)$

$\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}$

$\implies P(A\cap B)=P(A\cap B^C)$

We make use of the addition rule and complementary probabilities to rewrite this as

$P(A\cap B)=P(A\cap B^C)$

$\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)$

$\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)$

$\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]$

$\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]$

$\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C$

$\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)$

By the law of total probability,

$P(B)=P(A\cap B)+P(A^C\cap B)$

$\implies P(A^C\cap B)=P(B)-P(A\cap B)$

and substituting this into $(*)$ gives

$2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]$

$\implies P(B)=P(A\cup B)-P(A\cap B)$

$\implies P(B)=0.94-0.15=\boxed{0.79}$

8 0

3 years ago

Tell whether the ordered pair is a solution of y= -3x+9 <br><br> (2,3)

vodka [1.7K]

it is a solution :)))))

4 0

3 years ago

Read 2 more answers

Which of the following is a way that you would employ arithmetic in a real world situation?

otez555 [7]

Hey there! The answer is actually A. Choosing a grocery item based on a price per unit comparison.

I hope this helps you guys!

I have a screenshot attached as proof :)

4 0

3 years ago

Read 2 more answers