Solve for one of the variables in the first equation (in this case, I will solve for X):
X = Y + 3
Then use that value of X in the second equation to solve for Y:
X + 3Y = 9
(Y + 3) + 3Y = 9
4Y + 3 = 9
4Y = 6
Y = 1.5
Use the value of Y we just found in the X equation we created:
X = Y + 3
X = 1.5 + 3
X = 4.5
Therefore X = 4.5 or 9/2 and Y = 1.5 or 3/2.
Answer:
11
Step-by-step explanation:
5x+15+20=90
5x+35=90
-35 from both sides
5x=55
divide both sides by 5
x=11
1-39 ODD:
1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39
Now you'd add them all up.
1+3+5+7+9 = 25
11+13+15+17+19 = 75
21+23+25+27+29 = 125
31+33+35+37+39 = 175
Add the totals:
25+75+125+175
The answer would be 400.
Answer:
a) possible progressions are 5
b) the smallest and largest possible values of the first term are 16 and 82
Step-by-step explanation:
<u>Sum of terms:</u>
- Sₙ = n/2(a₁ + aₙ) = n/2(2a₁ + (n-1)d)
- S₂₀ = 20/2(2a₁ + 19d) = 10(2a₁ + 19d)
- 2020 = 10(2a₁ + 19d)
- 202 = 2a₁ + 19d
<u>In order a₁ to be an integer, d must be even number, so d = 2k</u>
- 202 = 2a₁ + 38k
- 101 = a₁ + 19k
<u>Possible values of k= 1,2,3,4,5</u>
- k = 1 ⇒ a₁ = 101 - 19 = 82
- k = 2 ⇒ a₁ = 101 - 38 = 63
- k = 3 ⇒ a₁ = 101 - 57 = 44
- k = 4 ⇒ a₁ = 101 - 76 = 25
- k = 5 ⇒ a₁ = 101 - 95 = 16
<u>As per above, </u>
- a) possible progressions are 5
- b) the smallest and largest possible values of the first term are 16 and 82
Answer:
For first lamp ; The resultant probability is 0.703
For both lamps; The resultant probability is 0.3614
Step-by-step explanation:
Let X be the lifetime hours of two bulbs
X∼exp(1/1400)
f(x)=1/1400e−1/1400x
P(X<x)=1−e−1/1400x
X∼exp(1/1400)
f(x)=1/1400 e−1/1400x
P(X<x)=1−e−1/1400x
The probability that both of the lamp bulbs fail within 1700 hours is calculated below,
P(X≤1700)=1−e−1/1400×1700
=1−e−1.21=0.703
The resultant probability is 0.703
Let Y be a lifetime of another lamp two bulbs
Then the Z = X + Y will follow gamma distribution that is,
X+Y=Z∼gamma(2,1/1400)
2λZ∼
X+Y=Z∼gamma(2,1/1400)
2λZ∼χ2α2
The probability that both of the lamp bulbs fail within a total of 1700 hours is calculated below,
P(Z≤1700)=P(1/700Z≤1.67)=
P(χ24≤1.67)=0.3614
The resultant probability is 0.3614
