36j + j = 37j
Explanation.
You could translate this sum into a normal sebtence: I have 36 apples and add 1 apple more.
The sum will then be 36 apples + apple = 37 apples.
Replacing the word (in this explanation 'apple') by a letter learns us 36a + a = 37a
Hence 36j + j = 37j
Answer:
0.80589
Step-by-step explanation:
So all of the numbers of correct answers less than 4 are 0,1,2,3
We need to calculate the probability for each separately and then add them together.
To find the probability we have to first find the combination. We know that there’s n=8 trials and that p=0.3. So 1-0.3 gives us 0.7.
The combination formula is: ! / (!(−)!)
So the n would always =8, and the r would be 0,1,2,3. So you would have to calculate it for 0,1,2,3 Separately. This can be done by hand or you can use a simple combinations calculator online.
For 0;
The combination is 1,
1 x 0.3^0 x 0.7^8-0 =
0.057648
For 1;
The combination is 8,
8 x 0.3^1 x 0.7^8-1 =
0.19765
For 2;
The combination is 28
28 x 0.3^2 x 0.7^8-2 =
0.296475
For 3;
The combination is 56
56 x 0.3^3 x 0.7^8-3 =
0.254122
All that’s left is to add these four numbers;
0.057647 + 0.19765 + 0.296475 + 0.254122 = 0.80589
Answer:
meter and then gallon
Step-by-step explanation:
i hope this helps :)
Answer:
There are no true solutions to the equation.
Step-by-step explanation:
<u><em>The correct equation is</em></u>
Solve for y
squared both sides
<em>Verify</em>
substitute the value of y in the original expression
----> is not true
therefore
There are no true solutions to the equation.
If you're using the app, try seeing this answer through your browser: brainly.com/question/2799412_______________
Let
(that is the range of the inverse sine function).
So,
![\mathsf{sin\,\theta=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\theta=x\qquad\quad(i)}](https://tex.z-dn.net/?f=%5Cmathsf%7Bsin%5C%2C%5Ctheta%3Dsin%5C%21%5Cleft%5Bsin%5E%7B-1%7D%28x%29%5Cright%5D%7D%5C%5C%5C%5C%20%5Cmathsf%7Bsin%5C%2C%5Ctheta%3Dx%5Cqquad%5Cquad%28i%29%7D)
Square both sides:
Since
then
is positive. So take the positive square root and you get
Then,
![\mathsf{tan\,\theta=\dfrac{sin\,\theta}{cos\,\theta}}\\\\\\ \mathsf{tan\,\theta=\dfrac{x}{\sqrt{1-x^2}}}\\\\\\\\ \therefore~~\mathsf{tan\!\left[sin^{-1}(x)\right]=\dfrac{x}{\sqrt{1-x^2}}\qquad\qquad -1\ \textless \ x\ \textless \ 1.}](https://tex.z-dn.net/?f=%5Cmathsf%7Btan%5C%2C%5Ctheta%3D%5Cdfrac%7Bsin%5C%2C%5Ctheta%7D%7Bcos%5C%2C%5Ctheta%7D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7Btan%5C%2C%5Ctheta%3D%5Cdfrac%7Bx%7D%7B%5Csqrt%7B1-x%5E2%7D%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C%20%5Ctherefore~~%5Cmathsf%7Btan%5C%21%5Cleft%5Bsin%5E%7B-1%7D%28x%29%5Cright%5D%3D%5Cdfrac%7Bx%7D%7B%5Csqrt%7B1-x%5E2%7D%7D%5Cqquad%5Cqquad%20-1%5C%20%5Ctextless%20%5C%20x%5C%20%5Ctextless%20%5C%201.%7D)
I hope this helps. =)
Tags: <em>inverse trigonometric function sin tan arcsin trigonometry</em>
