All the choices are correct.
_____
In our base-10 number system, moving a digit one place to the left multiplies its value by 10. Moving it one place to the right multiplies its value by 1/10.
You know already that 1,000 has 10 times the value of 100, and 1/10 the value of 10,000.
<h3>
Answer: C) 2 km west</h3>
Explanation:
With displacement, all we care about is the beginning and end. We don't care about the middle part(s) of the journey. So we'll take the straight line route from beginning to end when it comes to computing displacement.
We start at A(0,0) and end at B(-2,0). Going from A directly to B has us go 2 km west. Keep in mind that displacement is a vector, so you must include the direction along with the distance.
Answer:
Factors are (x-9)(x+4)
Step-by-step explanation:
X(3) - Y(2) = 14. So you need numbers that will multiply by 3 and a number that multiply by 2 that will equal 14.
The x can be 10 (equals 30) and then y can be 8 (equals 16).
30 subtracted by 16 equals 14.
ANSWER:
X = 10 Y = 16
3(10) - 2(8) = 14
Hope this helped. Have a great day. A thank you and (or) brainliest is always appreciated. Goodbye!
Answer:
The students should request an examination with 5 examiners.
Step-by-step explanation:
Let <em>X</em> denote the event that the student has an “on” day, and let <em>Y</em> denote the
denote the event that he passes the examination. Then,
The events (
) follows a Binomial distribution with probability of success 0.80 and the events (
) follows a Binomial distribution with probability of success 0.40.
It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,
Then,
⇒
Then,
Compute the probability that the students passes if request an examination with 3 examiners as follows:
![=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}](https://tex.z-dn.net/?f=%3D%5B%5Csum%5Climits%5E%7B3%7D_%7Bx%3D2%7D%7B%7B3%5Cchoose%20x%7D%280.80%29%5E%7Bx%7D%281-0.80%29%5E%7B3-x%7D%7D%5D%5Ctimes%5Cfrac%7B2%7D%7B3%7D%2B%5B%5Csum%5Climits%5E%7B3%7D_%7Bx%3D2%7D%7B%7B3%5Cchoose%20x%7D%280.40%29%5E%7B3%7D%281-0.40%29%5E%7B3-x%7D%7D%5D%5Ctimes%5Cfrac%7B1%7D%7B3%7D)
The probability that the students passes if request an examination with 3 examiners is 0.715.
Compute the probability that the students passes if request an examination with 5 examiners as follows:
![=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}](https://tex.z-dn.net/?f=%3D%5B%5Csum%5Climits%5E%7B5%7D_%7Bx%3D3%7D%7B%7B5%5Cchoose%20x%7D%280.80%29%5E%7Bx%7D%281-0.80%29%5E%7B5-x%7D%7D%5D%5Ctimes%5Cfrac%7B2%7D%7B3%7D%2B%5B%5Csum%5Climits%5E%7B5%7D_%7Bx%3D3%7D%7B%7B5%5Cchoose%20x%7D%280.40%29%5E%7Bx%7D%281-0.40%29%5E%7B5-x%7D%7D%5D%5Ctimes%5Cfrac%7B1%7D%7B3%7D)
The probability that the students passes if request an examination with 5 examiners is 0.734.
As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.
