Answer:
Step-by-step explanation:
The answer is 3 /8/33.
step by step explanation
First, we can write:
x
=
3
.
¯¯¯¯
24
Next, we can multiply each side by
100
giving:
100
x
=
324
.
¯¯¯¯
24
Then we can subtract each side of the first equation from each side of the second equation giving:
100
x
−
x
=
324
.
¯¯¯¯
24
−
3
.
¯¯¯¯
24
We can now solve for
x
as follows:
100
x
−
1
x
=
(
324
+
0
.
¯¯¯¯
24
)
−
(
3
+
0
.
¯¯¯¯
24
)
(
100
−
1
)
x
=
324
+
0
.
¯¯¯¯
24
−
3
−
0
.
¯¯¯¯
24
99
x
=
(
324
−
3
)
+
(
0
.
¯¯¯¯
24
−
0
.
¯¯¯¯
24
)
99
x
=
321
+
0
99
x
=
321
99
x
99
=
321
99
99
x
99
=
3
×
107
3
×
33
x
=
3
×
107
3
×
33
x
=
107
33
Now, we can convert this improper fraction to a mixed number:
x
=
107
33
=
99
+
8
33
=
99
33
+
8
33
=
3
+
8
33
=
3
8
33
3
.
¯¯¯¯
24
=
3
8
33
The answer is true because if you multiply 3•(3) is will give you 9 and if you multiply 10•(3) it will give you 30 which gives you both of the sides for the larger one
Answer:0.001001001001001
Step-by-step explanation: I have no idea. just put it in a calculator
Answer:
A and C
Step-by-step explanation:
To determine which events are equal, we explicitly define the elements in each set builder.
For event A
A={1.3}
for event B
B={x|x is a number on a die}
The possible numbers on a die are 1,2,3,4,5 and 6. Hence event B is computed as
B={1,2,3,4,5,6}
for event C
![C=[x|x^{2}-4x+3]\\solving x^{2}-4x+3\\x^{2}-4x+3=0\\x^{2}-3x-x+3=0\\x(x-3)-1(x-3)=0\\x=3 or x=1](https://tex.z-dn.net/?f=C%3D%5Bx%7Cx%5E%7B2%7D-4x%2B3%5D%5C%5Csolving%20%20x%5E%7B2%7D-4x%2B3%5C%5Cx%5E%7B2%7D-4x%2B3%3D0%5C%5Cx%5E%7B2%7D-3x-x%2B3%3D0%5C%5Cx%28x-3%29-1%28x-3%29%3D0%5C%5Cx%3D3%20or%20x%3D1)
Hence the set c is C={1,3}
and for the set D {x| x is the number of heads when six coins re tossed }
In the tossing a six coins it is possible not to have any head and it is possible to have head ranging from 1 to 6
Hence the set D can be expressed as
D={0,1,2,3,4,5,6}
In conclusion, when all the set are compared only set A and set C are equal
Answer:
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Step-by-step explanation: