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Maslowich
3 years ago
6

The probability of picking two black marbles from a box at random without replacement is 10/91 .

Mathematics
2 answers:
sasho [114]3 years ago
8 0
The probability of the 2 events are multiplied to get the result 10/91

so we have the relation 5/14 *  x  = 10/91

where x is the probability of drawing the second black

so x =  10/91 * 14/5 =  4/13   answer
Scorpion4ik [409]3 years ago
7 0

Answer:

4/13

<em />

<em>This is the correct answer! I just took the test! Good luck!!</em>

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Which statement is correct about the system of linear equations graphed below?
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D is the correct answer
8 0
3 years ago
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How many solutions are there to the system of equations? StartLayout enlarged left-brace 1st row 4 x minus 5 y = 5 2nd row negat
kkurt [141]

Answer:

Only one solution  x=\frac{5}{4},y=0

Step-by-step explanation:

4x-5y=5..............(1)\\0.08x+0.10y=0.10...........(2)

Multiply equation (2) by 50

4x-5y=5........(3)

Now from equation (1)+ equation(3)

8x=10\\x=\frac{10}{8}\\x=\frac{5}{4}

Put the value in equation (1)

5-5y=5\\5y=0\\y=0

hence only one solution exists\left(\frac{5}{4},0\right).

3 0
3 years ago
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Holly spends 7 3/5 hours in school each day. Her lunch is 1/2 hour long, and she spends 7/10 of a hour switching classes. The re
Ira Lisetskai [31]
<span>
y = 7 + 3/5
y = 35/5 + 3/5
y = 38/5
y = 2*(38/5)
y = 76/10
---
lunch time:
z = 1/2
z = 5*(1/2)
z = 5/10
---
time switching classes:
w = 7/10
---
y - 6x - z - w = 0
6x = y - z - w
x = (y - z - w)/6
x = (76/10 - 5/10 - 7/10)/6
x = (76 - 5 - 7)/(10*6)
x = (64)/(10*6)
x = (2*2*2*2*2*2)/(2*5*2*3)
x = (2*2*2*2)/(5*3)
x = 16/15

x = 1.0666666666
---
check:
y = 7 + 3/5
y = 7.6
z = 1/2
z = 0.5
w = 7/10
w = 0.7
y - 6x - z - w = 0
6x = y - z - w
x = (y - z - w)/6
x = (7.6 - 0.5 - 0.7)/6
x = 1.0666666666

answer:
 1.07 hours</span>
5 0
3 years ago
HELP
anyanavicka [17]
The axis of symmetry is x=2.
6 0
3 years ago
I can’t figure this out
jarptica [38.1K]

\dfrac{AB}{BC}=\dfrac{PQ}{QR}\\\\\dfrac{27}{45}=\dfrac{33}{QR}\\\\\dfrac{3}{5}=\dfrac{33}{QR}\\\\3QR=33\cdot5\\QR=11\cdot5\\QR=55

\dfrac{AB}{AC}=\dfrac{PQ}{PR}\\\\\dfrac{27}{36}=\dfrac{3}{PR}\\\\\dfrac{3}{4}=\dfrac{3}{PR}\\\\3PR=3\cdot4\\PR=4

3 0
3 years ago
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