Well, when you want to make 15 a percentage, You need to know of what, for an example, 20 percent of 100 is 20% percent because 20/100= 0.20 which equals 20%. What is 15 of? Fill in the blank. 15 percent of ___=? which would be 15/?
Answer:
8.4 in by 8.4 in
Step-by-step explanation:
The area of a square is the side length squared.
Let s be the side length:
s^2 = 70
s = \sqrt(70) = 8.4 in
Answer:
21.16
Step-by-step explanation:
Starting from the theory we have the following equation:
![fi*P(x](https://tex.z-dn.net/?f=fi%2AP%28x%3Cc-1%29%20%3D%200.99)
Using the data supplied in the exercise, we have subtracting the mean and dividing by the standard deviation:
![P( z \leq \frac{c-1-12}{3.5}) =0.99/fi](https://tex.z-dn.net/?f=P%28%20z%20%5Cleq%20%5Cfrac%7Bc-1-12%7D%7B3.5%7D%29%20%3D0.99%2Ffi)
solving for "c", knowing that fi is a tabulating value:
![\frac{c-13}{3.5}=0.99/fi\\\frac{c-13}{3.5}=2.33\\c-13=2.33*3.5\\c = 8.155 +13\\c = 21.155](https://tex.z-dn.net/?f=%5Cfrac%7Bc-13%7D%7B3.5%7D%3D0.99%2Ffi%5C%5C%5Cfrac%7Bc-13%7D%7B3.5%7D%3D2.33%5C%5Cc-13%3D2.33%2A3.5%5C%5Cc%20%3D%208.155%20%2B13%5C%5Cc%20%3D%2021.155)
therefore the value of c is equal to 21.16
I got 1/81. because u spun the spinner twice i multiplied the regions you are allowed to get by 2 which makes your denominator 18 then you can get the numbers 3 and 6 twice since the spinner is used twice then multiplied them together. so it would be 2/18 x 2/18 = 4/324 which equals 1/81
Answer:
a)
degrees
b) ![I_{T}(95,50) = -7.73](https://tex.z-dn.net/?f=I_%7BT%7D%2895%2C50%29%20%3D%20-7.73)
Step-by-step explanation:
An approximate formula for the heat index that is valid for (T ,H) near (90, 40) is:
![I(T,H) = 45.33 + 0.6845T + 5.758H - 0.00365T^{2} - 0.1565TH + 0.001HT^{2}](https://tex.z-dn.net/?f=I%28T%2CH%29%20%3D%2045.33%20%2B%200.6845T%20%2B%205.758H%20-%200.00365T%5E%7B2%7D%20-%200.1565TH%20%2B%200.001HT%5E%7B2%7D)
a) Calculate I at (T ,H) = (95, 50).
degrees
(b) Which partial derivative tells us the increase in I per degree increase in T when (T ,H) = (95, 50)? Calculate this partial derivative.
This is the partial derivative of I in function of T, that is
. So
![I(T,H) = 45.33 + 0.6845T + 5.758H - 0.00365T^{2} - 0.1565TH + 0.001HT^{2}](https://tex.z-dn.net/?f=I%28T%2CH%29%20%3D%2045.33%20%2B%200.6845T%20%2B%205.758H%20-%200.00365T%5E%7B2%7D%20-%200.1565TH%20%2B%200.001HT%5E%7B2%7D)
![I_{T}(T,H) = 0.6845 - 2*0.00365T - 0.1565H + 2*0.001H](https://tex.z-dn.net/?f=I_%7BT%7D%28T%2CH%29%20%3D%200.6845%20-%202%2A0.00365T%20-%200.1565H%20%2B%202%2A0.001H)
![I_{T}(95,50) = 0.6845 - 2*0.00365*(95) - 0.1565*(50) + 2*0.001(50) = -7.73](https://tex.z-dn.net/?f=I_%7BT%7D%2895%2C50%29%20%3D%200.6845%20-%202%2A0.00365%2A%2895%29%20-%200.1565%2A%2850%29%20%2B%202%2A0.001%2850%29%20%3D%20-7.73)