Answer:
D: 1/35
Step-by-step explanation:
First, you have to know how many marbles there are, and if you count them all up there are 15 marbles in total. Next, there are 3 blue marbles, so you would put the fraction as 3/15, or simplified to 1/5. Now, if you don't replace that blue marble, you now only have 14 marbles, and there are 2 yellow, which means you have a 2/14 chance of drawing a yellow marble, simplified to 1/7. But because there are two parts of the problem you are going to multiply 1/5 by 1/7 to get 1/35, which is your answer.
Answer:
Please check the explanation.
Step-by-step explanation:
To find the amount we use the formula:
Here:
A = total amount
P = principal or amount of money deposited,
r = annual interest rate
n = number of times compounded per year
t = time in years
Given
P=$2000
r=4.5%
n=4
t = 5 years
<em />
<u><em>Calculating compounded quarterly
</em></u>
After plugging in the values
Thus, If you deposit $2000 into an account paying 4.5% annual interest compounded quarterly, you will have $2501.50 after five years.
<u><em>Calculating compounded semi-annually</em></u>
n = 2
Thus, If you deposit $2000 into an account paying 4.5% annual interest compounded semi-annually, you will have $2,498.41 after five years.
180 = x + x+2 + x+1
180=3x+3
177= 3x
x = 59
Angle A = 59 +2 = 61 degrees
Supplementary angles is when 2 angles add up to 180 degrees.
We will name the smaller angle a and the bigger angle b.
therefore:
b - 172.6 = a
a + b = 180
Now we substitute:
a + b = 180
b - 172.6 + b = 180
2b - 172.6 = 180
2b = 352.6
b = 176.3
To find a:
a + b = 180
a + 176.3 = 180
a = 3.7
Therefore,
angle a = 3.7 degrees
angle b = 176.3 degrees.
Hope it helps and have a great day ahead
Answer:
The quadratic x2 − 5x + 6 factors as (x − 2)(x − 3). Hence the equation x2 − 5x + 6 = 0
has solutions x = 2 and x = 3.
Similarly we can factor the cubic x3 − 6x2 + 11x − 6 as (x − 1)(x − 2)(x − 3), which enables us to show that the solutions of x3 − 6x2 + 11x − 6 = 0 are x = 1, x = 2 or x = 3. In this module we will see how to arrive at this factorisation.
Polynomials in many respects behave like whole numbers or the integers. We can add, subtract and multiply two or more polynomials together to obtain another polynomial. Just as we can divide one whole number by another, producing a quotient and remainder, we can divide one polynomial by another and obtain a quotient and remainder, which are also polynomials.
A quadratic equation of the form ax2 + bx + c has either 0, 1 or 2 solutions, depending on whether the discriminant is negative, zero or positive. The number of solutions of the this equation assisted us in drawing the graph of the quadratic function y = ax2 + bx + c. Similarly, information about the roots of a polynomial equation enables us to give a rough sketch of the corresponding polynomial function.
As well as being intrinsically interesting objects, polynomials have important applications in the real world. One such application to error-correcting codes is discussed in the Appendix to this module.
