To start out, notice that you want the percent of voters that chose candidate A,
not the percent of the class that chose candidate A.
Your fraction should be "number that chose candidate A" out of "number of voters," which is the same thing as saying: "number that chose candidate A" divided by "number of voters"1) The numerator of the fraction should be the number of votes for candidate A, which is 11.
2) The denominator of the fraction should be the number of voters. You're told that "t<span>here were 11 votes for Candidate A and 15 votes for Candidate B," so there are:
</span>
![11 + 15 \: total \: votes](https://tex.z-dn.net/?f=11%20%2B%2015%20%5C%3A%20total%20%5C%3A%20votes)
3) Finally put parts 1 and 2 together into a fraction and multiply by 100 to get your percent. That is your final answer:
![\frac{11}{11+15} \times 100](https://tex.z-dn.net/?f=%20%5Cfrac%7B11%7D%7B11%2B15%7D%20%20%5Ctimes%20100)
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Answer: Top right choice, ![\frac{11}{11+15} \times 100](https://tex.z-dn.net/?f=%20%5Cfrac%7B11%7D%7B11%2B15%7D%20%5Ctimes%20100)
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Answer:
3x - y =13
Step-by-step explanation:
- y - y1 = m(x - x1)
- y - 2 = m(x - 5)
- distribute
- y - 2 = 3x - 15
- jump -2 across the equal sign and change the sign
- y = 3x - 15 + 2
- combine like terms
- y = 3x - 13
- use this equation to find the standard form
- y = 3x - 13
- jump the x and change the sign
- -3x + y = -13
- x must be positive, so change the sign of everything
- 3x - y = 13
Answer:
2i: 169.71
2ii: 0.17L
3a: 4×10⁻⁵
3b: 110011
Step-by-step explanation:
2i. The surface of the top and bottom of the tin is two times (top and bottom) π·r² = 2·π·3² = 18π cm².
The circumference of the circle is 2·π·r = 6π cm².
The area of the material connecting top and bottom is a rectangle of the tin height times the circumference: 6·6π = 36π cm².
This gives a total of 18π + 36π = 54π cm².
With π approximated by 22/7 the total surface area is 54*22/7 ≈ 169.71.
Notice how the calculation is simple by waiting until the very last moment to substitute π.
2ii. The volume is the area π·r² of the circle times the height of the tin: 9π*6 = 54π cm³ ≈ 169.71 cm³.
Since 1L = 1000 cm³ the volume is 0.16971 litres, which should be rounded to 0.17 L.
3a: If we rewrite P as 36 x 10⁻⁴ and realize that 36/2.25 = 16, then the fraction can be written as
16 x 10⁻⁴⁻⁶ = 16 x 10⁻¹⁰.
The square root of that is taking it to the power of 1/2, so (16x10⁻¹⁰)^0.5 = 4x10⁻⁵ = 0.00004
3b: 1111 1111 is 255 in decimal. 101 is 5 in decimal. 255/5 is 51 in decimal. 51 in binary is 110011.
The volume of the cylinder on the left is greater.
The volume of the cylinder on the right is greater.
The volumes are equal.