Answer:
for what you have you need to add one more $10 bill.
or make it total $120.25
good luck!!
Answer:
less than 1 1/2 gallons
Step-by-step explanation:
1/3 + 1/6 = 1/2, so the sum of the three cans is more than 1 by the difference between 1/5 and 1/6. That difference is 1/30 gallon. The sum is 1 1/30 gallons, which is less than 1 1/2 gallons.
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A suitable common denominator is 2·3·5 = 30. Then the sum of the fractions is ...
1/3 + 1/5 + 1/2
= 10/30 + 6/30 + 15/30
= 31/30 = 1 1/30 . . . . . less than 1 1/2
In decimal, 1/3 ≈ 0.333, 1/5 = 0.200, 1/2 = 0.500, so the sum is ...
0.333 +0.200 +0.500 = 1.033
which is less than 1.5.
The probability any one system works is 0.99
So the probability of any one system failing is 1-0.99 = 0.01, so basically a 1% chance of failure for any one system
Multiply out the value 0.01 with itself four times
0.01*0.01*0.01*0.01 = 0.000 000 01
I'm using spaces to make the number more readable
So the probability of all four systems failing is 0.00000001
Subtract this value from 1 to get
1 - 0.00000001 = 0.99999999
The answer is 0.99999999 which is what we'd expect. The probability of at least one of the systems working is very very close to 1 (aka 100%)
Answer:
Step-by-step explanation:
Solution by substitution method
3x-4y=8
and 18x-5y=10
Suppose,
3x-4y=8→(1)
and 18x-5y=10→(2)
Taking equation (1), we have
3x-4y=8
⇒3x=4y+8
⇒x=(
4y+8)/
3 →(3)
Putting x=
(4y+8
)/3 in equation (2), we get
18x-5y=10
18(
(4y+8)
/3) -5y=10
⇒24y+48-5y=10
⇒19y+48=10
⇒19y=10-48
⇒19y= -38
⇒y=-
38
/19
⇒y= -2→(4)
Now, Putting y=-2 in equation (3), we get
x=4y+8
x=
(4(-2)+8)
/3
⇒x=
(-8+8)/
3
⇒x=
0/
3
⇒x=0
∴x=0 and y= -2
Answer:
Step-by-step explanation:
The given piecewise function i
From the given function it is clear that function is divided at x=-1 and x=2. It means we check the discontinuity at x=-1 and x=2.
For x=-1,
LHL:
Since LHL ≠ f(-1), therefore the given function is discontinuous at x=-1.
For x=2,
LHL:
Since LHL ≠ f(2), therefore the given function is discontinuous at x=2.
Therefore, the correct option is A.
