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4 years ago

The mark-up for a 75 dollar golf club is 80 dollars. Find the percent of the mark-up.

1 answer:

4 0

So from 75 to 80 and 75 is 100%

80-75=5
5 is x% of 75
x%=percent markup
5=x% times 75
% means parts out of 100 so
x%=x/100
5=x/100 times 75
5=75x/100
75x/100=75/100 times x
75/100=3/4
5=3/4x
multiply both sides by 4/3
20/3=x
6.6666%
the percent mark up is 6.6666%

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3 years ago

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A store is selling 5 types of hard candies: cherry, strawberry, orange, lemon and pineapple. How many ways are there to choose:(

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Answer:

The number of ways of selecting of n items form r different items = $^{(n+r-1)}C_{(r-1)}$

A store is selling 5 types of hard candies: cherry, strawberry, orange, lemon and pineapple.

A) How many ways are there to choose 32 candies?

So, n = no. of items = 32

r = no. of types of items = 5

So, No. of ways to choose 32 candies = $^{(32+5-1)}C_{(5-1)}$

= $^{36}C_{4}$

= $\frac{36!}{4!(36-4)!}$

= $58905$

So, No. of ways to choose 32 candies is 58905

B)32 candies with at least a piece of each flavor?

Out of 32 you choose 5 candies of different types

So, Remaining candies = 32 - 5 = 27

So, No. of ways to choose 27 candies = $^{(27+5-1)}C_{(5-1)}$

= $^{31}C_{4}$

= $\frac{31!}{4!(31-4)!}$

= $31465$

So, No. of ways to choose 32 candies with at least a piece of each flavor is 31465

C) 32 candies with at least 4 cherry and at least 6 lemon?

So, you already choose 6+4= 10

So, remaining candies = 32-10 = 22

So, No. of ways to choose 22 candies = $^{(22+5-1)}C_{(5-1)}$

= $^{26}C_{4}$

= $\frac{26!}{4!(26-4)!}$

= $14950$

Hence No. of ways to choose 32 candies with at least 4 cherry and at least 6 lemon is 14950

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4 years ago

A news report says that 28% of high school students pack their lunch.

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Step-by-step explanation:

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4 years ago

Darren cut along a diagonal line across a sheet of paper to make a triangle. The paper was 10 inches long and 8 inches wide. Wha

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Answer:

$40 in^2$

Step-by-step explanation:

The initial dimensions of the paper are:

$L=10 in$ (length)

$W=8 in$ (width)

After the paper is cut along the diagonal, we remain with a right triangle, of which the length and the width corresponds to the base and the height.

For a triangle, the area is calculated as

$A=\frac{1}{2}bh$

where

b is the base

h is the height

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3 years ago

Maths functions <br> please help!

Vlad [161]

Answer:

$\textsf{1)} \quad f(x)=-x+3$

2) A = (3, 0) and C = (-3, 0)

$\textsf{3)} \quad g(x)=x^2-9$

4) AC = 6 units and OB = 9 units

Step-by-step explanation:

Given functions:

$\begin{cases}f(x)=mx+c\\g(x)=ax^2+b \end{cases}$

Given points:

H = (-1, 4)
T = (4, -1)

As points H and T lie on f(x), substitute the two points into the function to create two equations:

$\textsf{Equation 1}: \quad f(-1)=m(-1)+c=4 \implies -m+c=4$

$\textsf{Equation 2}: \quad f(4)=m(4)+c=-1 \implies 4m+c=-1$

Subtract the first equation from the second to eliminate c:

$\begin{array}{r l} 4m+c & = -1\\- \quad -m+c & = \phantom{))}4\\\cline{1-2}5m \phantom{))))}}& = -5}\end{aligned}$

Therefore m = -1.

Substitute the found value of m and one of the points into the function and solve for c:

$\implies f(4)=-1(4)+c=-1$

$\implies c=-1-(-4)=3$

Therefore the equation for function f(x) is:

$f(x)=-x+3$

Function f(x) crosses the x-axis at point A. Therefore, f(x) = 0 at point A.

To find the x-value of point A, set f(x) to zero and solve for x:

$\implies f(x)=0$

$\implies -x+3=0$

$\implies x=3$

Therefore, A = (3, 0).

As g(x) = ax² + b, its axis of symmetry is x = 0.

A parabola's axis of symmetry is the midpoint of its x-intercepts.

Therefore, if A = (3, 0) then C = (-3, 0).

Points on function g(x):

A = (3, 0)
G = (1, -8)

Substitute the points into the given function g(x) to create two equations:

$\textsf{Equation 1}: \quad g(3)=a(3)^2+b=0 \implies 9a+b=0$

$\textsf{Equation 2}: \quad g(1)=a(1)^2+b=-8 \implies a+b=-8$

Subtract the second equation from the first to eliminate b:

$\begin{array}{r l} 9a+b & = \phantom{))}0\\- \quad a+b & =-8\\\cline{1-2}8a \phantom{))))}}& = \phantom{))}8}\end{aligned}$

Therefore a = 1.

Substitute the found value of a and one of the points into the function and solve for b:

$\implies g(3)=1(3^2)+b=0$

$\implies 9+b=0\implies b=-9$

Therefore the equation for function g(x) is:

$g(x)=x^2-9$

The length AC is the difference between the x-values of points A and C.

$\implies x_A-x_C=3-(-3)=6$

Point B is the y-intercept of g(x), so when x = 0:

$\implies g(0)=(0)^2-9=-9$

Therefore, B = (0, -9).

The length OB is the difference between the y-values of the origin and point B.

$\implies y_O-y_B=0-(-9)=9$

Therefore, AC = 6 units and OB = 9 units

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2 years ago

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