The experimental probability so far is 19/34, since she made 19 good shots out of 34.
So, based on this results, and until new updates arrive, we can assume that she makes point with probability 19/34.
Answer:
Step-by-step explanation:
So in this example we'll be using the difference of squares which essentially states that: or another way to think of it would be: . So in this example you'll notice both terms are perfect squares. in fact x^n is a perfect square as long as n is even. This is because if it's even it can be split into two groups evenly for example, in this case we have x^8. so the square root is x^4 because you can split this up into (x * x * x * x) * (x * x * x * x) = x^8. Two groups with equal value multiplying to get x^8, that's what the square root is. So using these we can rewrite the equation as:
Now in this case you'll notice the degree is still even (it's 4) and the 4 is also a perfect square, and it's a difference of squares in one of the factors, so it can further be rewritten:
So completely factored form is:
I'm assuming that's considered completely factored but you can technically factor it further. While the identity difference of squares technically only applies to difference of squares, it can also be used on the sum of squares, but you need to use imaginary numbers. Because . and in this case a=x^2 and b=-4. So rewriting it as the difference of squares becomes: just something that might be useful in some cases.
Answer:
-x² + 16x - 60
Step-by-step explanation:
Given expression:
Step 1. <u>Insert</u> brackets to isolate the "-" from the (x - 8)².
⇒ -(x - 8)² + 4
⇒ -[(x - 8)²] + 4
Step 2. <u>Use</u> the formula (a - b)² = a² - 2ab + b²
⇒ -[(x - 8)²] + 4
⇒ -[x² - 2(x)(8) + 8²] + 4
Step 3. <u>Open</u> the parenthesis
⇒ -[x² - 2(x)(8) + 8²] + 4
⇒ -x² + 2(x)(8) - 8² + 4
Step 4. <u>Simplify</u> as needed
⇒ -x² + 16(x) - 64 + 4
⇒ -x² + 16x - 60
Answer:
a) The main idea to solve this exercise is to use the identity , where and are two square matrices.
Then, . Now, recall that [\det(Id) = \det(P)\det(P^{-1})[/tex], where stands for the identity matrix. But , thus and are reciprocal to each other.
Hence,
b) Let us write and . Then
But the product of two diagonal matrices is commutative, so , from where the statement readily follows.
Step-by-step explanation: