4 is a coefficient and X and Y are variables so the -110 is the constant so there for the answer is D
Answer:
see below
Step-by-step explanation:
The area of a circle is given by
A = pi r^2 where r is the radius
A = pi (6.5)^2
A =42.25 pi
If we approximate pi by 3.14
A = 42.25 * 3.14 =132.665
If we use the pi button
A = 132.7322896
Answer:
1) A) Increases
2) C) Stays the same
Step-by-step explanation:
1) A) Increases (94 to 95)
2) C) Stays the same (95)
Answer:
12.
Step-by-step explanation:
This equation seems to want you to use the Order of Operations, a.k.a PEMDAS (which stands for Parentheses, Exponents, Multiplication, Division, Addition, Subtraction). When you do a problem like this, you use the Order of Operations to figure out the order in which you'd solve your problem.
You would first solve what is in parentheses, since that is the first letter in PEMDAS. 18-6x2 is what we have in our parentheses, but it's too confusing to be solved yet. The next letter is E, a.k.a exponents, but there are no exponents in this equation, so we move on. Next is M, Multiplication. We would multiply 6 by 2, which gives us 12. Moving on, we have D, Division. We would do 36/6, which is 6.
Our 5th letter is A, Addition. We would have to add our 6 to (18-12), but that couldn't work, since our 18-12 is still in parentheses, and those come first. So, we would skip to the last letter, S, Subtraction. 18-12=6, and since we solved all the problems in the parentheses, we remove them, and add both 6's together, giving us 12.
Answer:
P(X ≥ 1) = 0.50
Step-by-step explanation:
Given that:
The word "supercalifragilisticexpialidocious" has 34 letters in which 'i' appears 7 times in the word.
Then; the probability of success = 7/34 = 0.20588
Using Binomial distribution to determine the probability; we have:
where;
x = 0,1,2,...n and 0 < β < 1
and x represents the number of successes.
However; since the letter is drawn thrice; the probability that the letter "i" is drawn at least once can be computed as:
P(X ≥ 1) = 1 - P(X< 1)
P(X ≥ 1) = 1 - P(X =0)
![P(X \ge 1) = 1 - \bigg [ {^3C__0} (0.21)^0 (1-0.21)^{3-0} \bigg]](https://tex.z-dn.net/?f=P%28X%20%5Cge%201%29%20%3D%20%201%20-%20%5Cbigg%20%5B%20%7B%5E3C__0%7D%20%280.21%29%5E0%20%281-0.21%29%5E%7B3-0%7D%20%5Cbigg%5D)
![P(X \ge 1) = 1 - \bigg [ 1 \times 1 (0.79)^{3} \bigg]](https://tex.z-dn.net/?f=P%28X%20%5Cge%201%29%20%3D%20%201%20-%20%5Cbigg%20%5B%201%20%5Ctimes%201%20%280.79%29%5E%7B3%7D%20%5Cbigg%5D)
P(X ≥ 1) = 1 - 0.50
P(X ≥ 1) = 0.50
