Answer:
A
Step-by-step explanation:
Assuming you require the distance between the 2 points.
To calculate the distance d use the distance formula
d = ![\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }](https://tex.z-dn.net/?f=%5Csqrt%7B%28x_%7B2%7D-x_%7B1%7D%29%5E2%2B%28y_%7B2%7D-y_%7B1%7D%29%5E2%20%20%20%20%7D)
with (x₁, y₁ ) = (- 3, - 8) and (x₂, y₂ ) = (- 2, - 3)
d = ![\sqrt{(-2+3)^2+(-3+8)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%28-2%2B3%29%5E2%2B%28-3%2B8%29%5E2%7D)
= ![\sqrt{1^2+5^2}](https://tex.z-dn.net/?f=%5Csqrt%7B1%5E2%2B5%5E2%7D)
= ![\sqrt{1+25}](https://tex.z-dn.net/?f=%5Csqrt%7B1%2B25%7D)
=
→ A
Answer:
21. C
22. D I got 263.89
23. I'm not sure about this one
24. C
25. D
Step-by-step explanation:
I hope the ones i answered were correct if not i am sorry
Given mean = 0 C and standard deviation = 1.00
To find probability that a random selected thermometer read less than 0.53, we need to find z-value corresponding to 0.53 first.
z= ![\frac{x-mean}{standard deviation} = \frac{0.53-0}{1.00} = 0.53](https://tex.z-dn.net/?f=%5Cfrac%7Bx-mean%7D%7Bstandard%20deviation%7D%20%20%3D%20%5Cfrac%7B0.53-0%7D%7B1.00%7D%20%3D%200.53)
So, P(x<0.53) = P(z<0.53) = 0.701944
Similarly P(x>-1.11)=P(z>-1.11) = 1-P(z<-1.11) = 0.8665
For finding probability for in between values, we have to subtract smaller one from larger one.
P(1.00<x<2.25) = P(1.00<z<2.25) = P(z<2.25)- P(z<1.00) = 0.9878 - 0.8413 = 0.1465
P(x>1.71) = P(z>1.71) = 1-P(z<1.71) = 1-0.9564 = 0.0436
P(x<-0.23 or x>0.23) = P(z<-0.23 or z>0.23) =P(z<-0.23)+P(z>0.23) = 0.409+0.409 = 0.918
1. V = r² π hV = 3² π · 10 = 90 π in³Answer: C.2. r = 18/2 = 9 yd, h = 3 ydV = 9² π · 3 = 243 π yd³3. r = 46.25 / 2 = 23.125 cmh = 18.5 cmV = 23.125² · 3.14 · 18.5 = 31,064.53 cm³4. h = 1934, d = 1934 · 125 = 241,750 , r = 241.750 / 2 = 120,875 V = 120,875² · 3.14 · 1934 = = 14,610,765,625 · 3.14 · 1934 = = 88,727,673,056,875