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katrin2010 [14]
3 years ago
13

Question below. Please Help!

Mathematics
2 answers:
chubhunter [2.5K]3 years ago
7 0

Answer:

Look to the top left square. look at -2, and go up by 3. Plot your point there.

Step-by-step explanation:

SpyIntel [72]3 years ago
5 0

Answer:

LEft bottom box

Step-by-step explanation:

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Which two inequalities can be used to find the solution to this absolute value inequality?
yanalaym [24]

Answer: 3(x+4)<-12 and x+4<4

Step-by-step explanation:

3|x+4| - 5 < 7

5 0
2 years ago
One of ten different prizes was randomly put into each box of a cereal. If a family decided to buy this cereal until it obtained
Reptile [31]

Answer:

29.29

Step-by-step explanation:

The first trial among a distribution of independent trials with a constant success probability follows a geometric probability.

The expected value of a negative binomial is the reciprocal of its success probability <em>p.</em>

<em>E(X)=</em>1/<em>p</em>

At the first draw, we only selected one of the 10 prizes and so 1 draw gives a success

<em>E(X)=</em>1/<em>p</em>

<em>E(X1)</em>=1

After the first prize is drawn, we are interested in the first time (success), that we draw another prize that is different (<em>p</em>=9/10)

E(X2)=1/9÷10

E(X2)= 10/9

After the first prizes have being drawn, we now interested in the first time success that we draw a different prize from the first 2

E(X3)=1/8÷10

E(X3)=10/8

E(X3)=5/4

After the first three prizes have been drawn, we are now interested in the success of a first time draw for the 4th different prize

E(X4)=1/7÷10

E(X4)=10/7

Fifth different prize

E(X5)=1/6÷10

E(X5)=10/6

E(X5)=5/3

Sixth different prize

E(X6)=1/5÷10

E(X6)=10/5

E(X6)=2

Seventh different prize

E(X7)=1/4÷10

E(X7)=10/4

E(X7)=5/2

Eighth different prize

E(X8)=1/3÷10

E(X8)=10/3

Ninth different prize

E(X9)=1/2÷10

E(X9)=10/2

E(X9)=5

After the 9 different prizes have been drawn, we are interested in the first time we will draw the tenth different prize

E(X10)=1/1÷10

E(X10)=10

Add up all corresponding values;

<em>E(X)=</em>E(X1)+E(X2)+E(X3)+E(X4)+E(X5)+E(X6)+E(X7)+E(X8)+E(X9)+E(X10)

<em>E(X)=</em>1+10/9+5/4+10/7+5/3+2+5/2+10/3+5+10

<em>E(X)</em>= 29.29

Note: Those values in <em>E(X), </em>should be written the way it will in standard algebra ie they should be small.

4 0
3 years ago
10 = 3x – 8<br><br> what is this
Margaret [11]

Answer:

<u><em>Answer is below</em></u>

Step-by-step explanation:

<u><em>10=3x-8</em></u>

<u><em>always remember to flip the equation</em></u>

<u><em>3x-8=10</em></u>

<u><em>Add 8 to both sides</em></u>

<u><em>3x-8(+8)=10(+8)</em></u>

<u><em>3x=18</em></u>

<u><em>Divide both sides by 3</em></u>

<u><em>3x/3=18/3</em></u>

<u><em>x=6</em></u>

8 0
3 years ago
Read 2 more answers
System of equations, special case infinitely many solutions?????
Mashcka [7]

One possible system is

1x + 3y = 4

2x + 6y = 8

Note how 2 is twice as large as 1, 6 is twice as large as 3, and 8 is twice as large as 4.

In other words, the second equation is the result of multiplying both sides of the first equation by 2.

1x+3y = 4

2*(1x+3y) = 2*4

2x+6y = 8

Effectively the two equations in bold are the same which produces the same line. The two lines overlap perfectly to intersect infinitely many times. An intersection is a solution.

4 0
3 years ago
Solve this inequality: 3p – 6 &gt; 21.
Wittaler [7]
<span>To solve for p:
3p – 6 > 21
</span><span>3p - 6 + 6 > 21 + 6 (add 6 to both sides)
</span><span>3p > 27
3p/3 </span><span>> 27/3 (divide both sides by 3 to get p by itself)
</span><span>p > 9</span>
4 0
3 years ago
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