Answer:
![\frac{4}{9}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B9%7D)
Step-by-step explanation:
The equation would be
![\frac{1}{3} +\frac{1}{9} =total](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%20%2B%5Cfrac%7B1%7D%7B9%7D%20%3Dtotal)
First, change the denominator to a common multiple.
The LCM (least common multiple) of 3 and 9 is 9.
![\frac{3}{9} +\frac{1}{9} =total](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B9%7D%20%2B%5Cfrac%7B1%7D%7B9%7D%20%3Dtotal)
Add to solve.
![\frac{4}{9}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B9%7D)
I hope this helps!
Answer:
7 percent
Step-by-step explanation:
We have the following information from the question:
The probability that a student chooses art as an elective is 15 percent: ![P(A)=15\%](https://tex.z-dn.net/?f=P%28A%29%3D15%5C%25)
The probability that a student chooses art or drama is 20 percent: ![P(A\cup D)=20\%](https://tex.z-dn.net/?f=P%28A%5Ccup%20D%29%3D20%5C%25)
The probability that a student chooses both art and drama is 2 percent:
![P(A\cap D)=2\%](https://tex.z-dn.net/?f=P%28A%5Ccap%20D%29%3D2%5C%25)
Recall that:
![P(A\cup D)=P(A)+P(D)-P(A\cap D)](https://tex.z-dn.net/?f=P%28A%5Ccup%20D%29%3DP%28A%29%2BP%28D%29-P%28A%5Ccap%20D%29)
We plug in the various probabilities to get:
![20\%=15\%+P(D)-2\%](https://tex.z-dn.net/?f=20%5C%25%3D15%5C%25%2BP%28D%29-2%5C%25)
![20\%-15\%+2\%=P(D)](https://tex.z-dn.net/?f=20%5C%25-15%5C%25%2B2%5C%25%3DP%28D%29)
![7\%=P(D)](https://tex.z-dn.net/?f=7%5C%25%3DP%28D%29)
![\therefore P(D)=7\%](https://tex.z-dn.net/?f=%5Ctherefore%20P%28D%29%3D7%5C%25)
Hence the probability that a student chooses drama as an elective is 7 percent
Answer:
775 milliliters of juice left
Step-by-step explanation:
1000mL in a liter
1000-275= 775
There is one solution because the lines intersect only once.
Recall the double angle identity for cosine:
![\cos(2x) = \cos^2(x) - \sin^2(x) = 1 - 2 \sin^2(x)](https://tex.z-dn.net/?f=%5Ccos%282x%29%20%3D%20%5Ccos%5E2%28x%29%20-%20%5Csin%5E2%28x%29%20%3D%201%20-%202%20%5Csin%5E2%28x%29)
It follows that
![\sin^2(x) = \dfrac{1 - \cos(2x)}2 \implies \sin(x) = \pm \sqrt{\dfrac{1-\cos(2x)}2} \implies \csc(x) = \pm \sqrt{\dfrac2{1-\cos(2x)}}](https://tex.z-dn.net/?f=%5Csin%5E2%28x%29%20%3D%20%5Cdfrac%7B1%20-%20%5Ccos%282x%29%7D2%20%5Cimplies%20%5Csin%28x%29%20%3D%20%5Cpm%20%5Csqrt%7B%5Cdfrac%7B1-%5Ccos%282x%29%7D2%7D%20%5Cimplies%20%5Ccsc%28x%29%20%3D%20%5Cpm%20%5Csqrt%7B%5Cdfrac2%7B1-%5Ccos%282x%29%7D%7D)
Since 0° < 22° < 90°, we know that sin(22°) must be positive, so csc(22°) is also positive. Let x = 22°; then the closest answer would be C,
![\csc(22^\circ) = \sqrt{\dfrac2{1-\cos(44^\circ)}} = \sqrt{\dfrac2{1-\frac5{13}}} = \dfrac{\sqrt{13}}2](https://tex.z-dn.net/?f=%5Ccsc%2822%5E%5Ccirc%29%20%3D%20%5Csqrt%7B%5Cdfrac2%7B1-%5Ccos%2844%5E%5Ccirc%29%7D%7D%20%3D%20%5Csqrt%7B%5Cdfrac2%7B1-%5Cfrac5%7B13%7D%7D%7D%20%3D%20%5Cdfrac%7B%5Csqrt%7B13%7D%7D2)
but the problem is that none of these claims are true; cot(32°) ≠ 4/3, cos(44°) ≠ 5/13, and csc(22°) ≠ √13/2...