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Sliva [168]
3 years ago
11

Find two numbers whose sum is 58 and whose difference is 16 Can anyone help me out

Mathematics
1 answer:
White raven [17]3 years ago
7 0
X, y - the numbers

x+y=58 \\
\underline{x-y=16} \\
x+x+y-y=58+16 \\
2x=74 \\
x=37 \\ \\
x+y=58 \\
37+y=58 \\
y=21

The two numbers are 37 and 21.
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in a standard casino dice game the roller wins on the first roll if he rolls a sum of 7 or 11. what is the probability of winnin
Alborosie

<u>Answer-</u>

<em>The probability of winning on the first roll is </em><em>0.22</em>

<u>Solution-</u>

As in the game of casino, two dice are rolled simultaneously.

So the sample space would be,

|S|=6^2=36

Let E be the event such that the sum of two numbers are 7, so

E = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}

|E|=56

\therefore P(E)=\dfrac{|E|}{|S|}=\dfrac{6}{36}

Let F be the event such that the sum of two numbers are 11, so

F = {(6,5), (5,6)}

|F|=2

\therefore P(F)=\dfrac{|F|}{|S|}=\dfrac{2}{36}

Now,

P(\text{sum is 7 or 11)}=P(E\ \cup\ F)=P(E)+P(F)=\dfrac{6}{36}+\dfrac{2}{36}=\dfrac{8}{36}=\dfrac{2}{9}=0.22

8 0
4 years ago
Explain why there must be at least two lines on any given plane
Lesechka [4]
BECAUSE THERE IS A Y AXIS AND AN X AXIS THOSE R THE 2 GIVIN LINES
6 0
4 years ago
Alex works as a nanny. Her regular hourly wage is $10.25. If she regularly works 28 hours per week, what is her regular weekly p
marishachu [46]

Answer:

She normally gets $287 a week

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
So like I just need help doing this ty​
Aleksandr-060686 [28]

Answer:

x=7

Step-by-step explanation:

9x -5 = 58 since they are vertical angles and vertical angles are equal

9x-5 = 58

Add 5 to each side

9x-5+5 = 58+5

9x = 63

Divide each side by 9

9x/9 = 63/9

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7 0
4 years ago
If 10-10ˣ=9.9, find x
madam [21]

we are given

10-10^x=9.9

we can move 10 on right side

-10^x=9.9-10

-10^x=-0.1

we cancel -1

10^x=0.1

we can rewrite right side

10^x=10^{-1}

we can compare exponents

and we get

x=-1.............Answer

3 0
3 years ago
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