For this case we have:
Be a function of the form ![y = f (x)](https://tex.z-dn.net/?f=y%20%3D%20f%20%28x%29)
Where:
![f (x) = \sqrt {x-5}](https://tex.z-dn.net/?f=f%20%28x%29%20%3D%20%5Csqrt%20%7Bx-5%7D)
If we want to find f (-2), we substitute
, then:
![f (-2) = \sqrt {-2-5}\\f (-2) = \sqrt {-7}](https://tex.z-dn.net/?f=f%20%28-2%29%20%3D%20%5Csqrt%20%7B-2-5%7D%5C%5Cf%20%28-2%29%20%3D%20%5Csqrt%20%7B-7%7D)
Since we have a negative root, the result will be given by complex numbers. By definition:
![\sqrt {-1} = i](https://tex.z-dn.net/?f=%5Csqrt%20%7B-1%7D%20%3D%20i)
So:
![f (-2) = \sqrt {7} i](https://tex.z-dn.net/?f=f%20%28-2%29%20%3D%20%5Csqrt%20%7B7%7D%20i)
Answer:
![f (-2) = \sqrt {7} i](https://tex.z-dn.net/?f=f%20%28-2%29%20%3D%20%5Csqrt%20%7B7%7D%20i)
Step-by-step explanation:
1. P(light OR domestic) = P(light) + P(domestic) − P(light AND domestic)
P(light OR domestic) = 0.62 + 0.70 − 0.55
P(light OR domestic) = 0.77
2. P(light AND not domestic) = P(light) − P(light AND domestic)
P(light AND not domestic) = 0.62 − 0.55
P(light AND not domestic) = 0.07
3. P(light GIVEN not domestic) = P(light AND not domestic) / P(not domestic)
P(light GIVEN not domestic) = 0.07 / (1 − 0.70)
P(light GIVEN not domestic) = 0.233
4. Two events are independent if P(A) × P(B) = P(A and B).
P(light) × P(domestic) = 0.62 × 0.70 = 0.434
P(light AND domestic) = 0.55
Therefore, the type and location are not independent.
Answer:
52
Step-by-step explanation:
(-6)x(-6)=36
(4)x(4)=16
36+16=52
Answer:
No. The new height of the water is less than the height of the glass(6.33 cm<10 cm)
Step-by-step explanation:
-For the water in the glass to overflow, the volume of the inserted solid must be greater than the volume of the empty space or the ensuing height of water >height of glass.
#Volume of the golf ball:
![V=\frac{4}{3}\pi r^3\\\\=\frac{4}{3}\pi \times 4^3\\\\\approx 268.08\ cm^3](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E3%5C%5C%5C%5C%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%5Ctimes%204%5E3%5C%5C%5C%5C%5Capprox%20268.08%5C%20cm%5E3)
#The volume of the water in the glass:
![V=\pi r^2 h\\\\=\pi \times 4^2\times 10\\\\\approx 50.27\ cm^2](https://tex.z-dn.net/?f=V%3D%5Cpi%20r%5E2%20h%5C%5C%5C%5C%3D%5Cpi%20%5Ctimes%204%5E2%5Ctimes%2010%5C%5C%5C%5C%5Capprox%2050.27%5C%20cm%5E2)
We then equate the two volumes to the glass' volume to determine the new height of the water:
![V=\pi r^2h\\\\(206.08+50.27)=\pi r^2 h\\\\h=318.35/(\pi \times 4^2)\\\\=6.33\ cm](https://tex.z-dn.net/?f=V%3D%5Cpi%20r%5E2h%5C%5C%5C%5C%28206.08%2B50.27%29%3D%5Cpi%20r%5E2%20h%5C%5C%5C%5Ch%3D318.35%2F%28%5Cpi%20%5Ctimes%204%5E2%29%5C%5C%5C%5C%3D6.33%5C%20cm)
Hence, the glass will not overflow since the new height of the water is less than the height of the glass(6.33 cm<10cm).