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horrorfan [7]
4 years ago
14

Please Help! (no rush)

Mathematics
2 answers:
Anon25 [30]4 years ago
3 0

Answer:

The answer to your question is below

Step-by-step explanation:

Formula  an expression that uses variables to state a rule

Constant   a number; a term containing no variables

Variable  a letter or symbol used to represent an unknown number

Term   a number or a variable, or the product of a number and variable(s)

Expression   one term or multiple terms connected by an addition or subtraction sign

Mama L [17]4 years ago
3 0

constant; a number; a term containing no variables

expression; one term or multiple terms connected by an addition or subtraction sign

term; an expression that uses variables to state a rule

formula;a number or a variable, or the product of a number and variable(s)

variable; a letter or symbol used to represent an unknown number


sorry if it’s wrong .
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The answer is (4) because you multiply the like terms in the equation. 5*7=35 and 3i*2i=6i.

Answer is (4)
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3 years ago
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Ksivusya [100]

Answer:

31

Step-by-step explanation:

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-5/4 multiply by 1/3
andriy [413]

The answer to this problem is 5/12.

4 0
4 years ago
Suppose you pay a dollar to roll two dice. if you roll 5 or a 6 you Get your dollar back +2 more just like it the goal will be t
LiRa [457]

Answer:

(a)$67

(b)You are expected to win 56 Times

(c)You are expected to lose 44 Times

Step-by-step explanation:

The sample space for the event of rolling two dice is presented below

(1,1), (2,1), (3,1), (4,1), (5,1), (6,1)\\(1,2), (2,2), (3,2), (4,2), (5,2), (6,2)\\(1,3), (2,3), (3,3), (4,3), (5,3), (6,3)\\(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)

Total number of outcomes =36

The event of rolling a 5 or a 6 are:

(5,1), (6,1)\\ (5,2), (6,2)\\( (5,3), (6,3)\\ (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)

Number of outcomes =20

Therefore:

P(rolling a 5 or a 6)  =\dfrac{20}{36}

The probability distribution of this event is given as follows.

\left|\begin{array}{c|c|c}$Amount Won(x)&-\$1&\$2\\&\\P(x)&\dfrac{16}{36}&\dfrac{20}{36}\end{array}\right|

First, we determine the expected Value of this event.

Expected Value

=(-\$1\times \frac{16}{36})+ (\$2\times \frac{20}{36})\\=\$0.67

Therefore, if the game is played 100 times,

Expected Profit =$0.67 X 100 =$67

If you play the game 100 times, you can expect to win $67.

(b)

Probability of Winning  =\dfrac{20}{36}

If the game is played 100 times

Number of times expected to win

=\dfrac{20}{36} \times 100\\=56$ times

Therefore, number of times expected to loose

= 100-56

=44 times

8 0
4 years ago
I have one last question I need help with.
klio [65]
Just the first one, r is 4x+2, and h is 5x+4, plug them into the formula pls
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