Since they are already in order you know the 22 and the 61 are the upper and lower extremes. To find the median you find the number that is in the middle which is 42. To find the lower quartile find the middle number starting from the first number and the number to the left of the median (22 and 36 in this case). The loser quartile is 25. To find the upper quartile, fine the middle number between the number to the right of the median and the last number (44 and 61 in this case.) the upper quartile is 57 you then just plot it and graph it
The work is attached
Answer:
vb hyjv bnhyn g
Step-by-step explanation:
Answer:
cos q = 3/5
Step-by-step explanation:
Standard position means the vertex (point or corner of the angle) is at (0,0) and one side of the angle is glued to the positive x-axis (facts, but not technical math terms) See image. Special triangles have all three sides nice and clean with whole number lengths, we call these Pythagorean triples. 3-4-5 is your most basic Pythagorean triple. So we don't even have to calculate the hypotenuse, see image. Now the triangle shown is easy to work with, using entry-level trig...cos = ADJ/HYP. So we get 3/5=.6 BUUuuuut, the angle q in the original problem is actually the giant angle, marked in yellow (see image) and we're in the fourth quadrant which means there's negative numbers all over the place. So just to be sure the answer is .6 and not -.6 Check your signs. One trick to remember is the ASTC markings in the quadrants. I use All Students Take Calculus, but what it means is in the first quadrant All the trig functions are positive. Only Sine (and fam) are positive in the 2nd quadrant. Tan (and fam) in the 3rd and Cos and fam in the 4th quadrant. It's a good quick check.
cos q = 3/5 OR cos q = .6
<span>-12 = pico(p)
So 10^-12 L = 1 pL
Use following to find the count of 10^-12 L are there in 2.3×10^-10 L
2.3×10^-10 / (10^-12) = 2.3×10^2 = 230 pL
-6 = micro (µ)
4.7×10^-6 g = 4.7 µg
1.85×10^-12 m = 1.85 pm
6 = Mega (M)
16.7×10^6 s = 16.7 Ms
So the answer is 16.7 Ms</span>
