Answer:
The value of A is 5
Step-by-step explanation:
- The number is divisible by 3 if the sum of its digits is a number
divisible by 3
- Ex: 126 is divisible by 3 because the sum of its digits = 1 + 2 + 3 = 6
and 6 is divisible by 3
- The number is divisible by 5 if its ones digit is zero or 5
- Ex: 675 is divisible by 5 because its ones digit is 5
890 is divisible by 5 because its ones digit is 0
- We are looking for the value of A in the 4-digit number 3A5A which
makes the number divisible by both 3 and 5
∵ A is in the ones position
∴ A must be zero or 5
- Let us try A = 0
∵ A = 0
∴ The number is 3050
∵ The sum of the digits of the number = 3 + 0 + 5 + 0 = 8
∵ 8 is not divisible by 3
∴ 3050 is not divisible by both 3 and 5
∴ A can not be zero
- Let us try A = 5
∵ A = 5
∴ The number is 3555
∵ The sum of the digits of the number = 3 + 5 + 5 + 5 = 18
∵ 18 is divisible by 3
∴ 3555 is divisible by both 3 and 5
∴ A must be equal 5
* <em>The value of A is 5</em>
There is only one solution for the equation 4z + 2(z -4) = 3z + 11 because the exponent for the power of z is 1.
<h3>What is an equation?</h3>
In mathematics, an equation is a formula that expresses the equality of two expressions, by connecting them with the equals sign =.
<h3>What is the Solution?</h3>
A solution is any value of a variable that makes the specified equation true.
According to the given information:
4z + 2(z-4)= 3z+11
Solve the equation,
4z+2z-8=3z+11
6z-3z=11+8
3z =19
z=
Hence,
Number of solution that can be found for the equation 4z + 2(z-4)= 3z+11 is option(2) one
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3x + 1 = y
2x + 3y = 14
To solve this system of equations, we are going to use the substitution method. Substitution the equation where the variable is isolated into the second equation. In this system of equations, y is isolated, so we will replace y in the second equation with 3x + 1.
2x + 3y = 14
2x + 3(3x + 1) = 14
2x + 9x + 3 = 14
We will add the like terms and subtract 3 from both sides of the equation.
11x + 3 = 14
11x = 11
x = 1
In this system of equations, x is equal to 1. Now we will go back and solve for y, plugging in 1 for x.
3(1) + 1 = y
2(1) + 3y = 14
3 + 1 = y
2 + 3y = 14
4 = y
3y = 2
4 = y
4 = y
The solution to this system of equations is (1, 4).
Answer:,n ,
Step-by-step explanation:
Let
be the total amount of money paid by any given set of passengers. If there are
passengers in a car, then the driver must pay a toll of
.
Then
has first moment (equal to the mean)
![E[Y]=E[0.5X+3]=0.5E[X]+3E[1]=0.5\mu_X+3=\boxed{4.35}](https://tex.z-dn.net/?f=E%5BY%5D%3DE%5B0.5X%2B3%5D%3D0.5E%5BX%5D%2B3E%5B1%5D%3D0.5%5Cmu_X%2B3%3D%5Cboxed%7B4.35%7D)
and second moment
![E[Y^2]=E[0.25X^2+3X+9]=0.25E[X^2]+3E[X]+9E[1]=0.25E[X^2]+3\mu_X+9](https://tex.z-dn.net/?f=E%5BY%5E2%5D%3DE%5B0.25X%5E2%2B3X%2B9%5D%3D0.25E%5BX%5E2%5D%2B3E%5BX%5D%2B9E%5B1%5D%3D0.25E%5BX%5E2%5D%2B3%5Cmu_X%2B9)
Recall that the variance is the difference between the first two moments:
![\mathrm{Var}[X]=E[X^2]-E[X]^2\implies E[X^2]={\sigma^2}_X+{\mu_X}^2](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BX%5D%3DE%5BX%5E2%5D-E%5BX%5D%5E2%5Cimplies%20E%5BX%5E2%5D%3D%7B%5Csigma%5E2%7D_X%2B%7B%5Cmu_X%7D%5E2)
![\implies E[Y^2]=0.25({\sigma^2}_X+{\mu_X}^2)+3\mu_X+9\approx19.22](https://tex.z-dn.net/?f=%5Cimplies%20E%5BY%5E2%5D%3D0.25%28%7B%5Csigma%5E2%7D_X%2B%7B%5Cmu_X%7D%5E2%29%2B3%5Cmu_X%2B9%5Capprox19.22)
![\implies\mathrm{Var}[Y]=E[Y^2]-E[Y]^2=\boxed{0.3}](https://tex.z-dn.net/?f=%5Cimplies%5Cmathrm%7BVar%7D%5BY%5D%3DE%5BY%5E2%5D-E%5BY%5D%5E2%3D%5Cboxed%7B0.3%7D)