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Dmitry [639]
3 years ago
14

you have 3.75 to spend at a vending machine. You want to buy at least three health bars. Regualr health bars cost 0.75 each and

strawberry health bar cost 1.25 each. Write a sysem of equtions that represents the situation?
Mathematics
1 answer:
DedPeter [7]3 years ago
4 0

Answer:

Let x be the number of regular health bars you buy and y the number of strawberry health bars you buy. Then:

0.75x+1.25y=3.75

x+y>=3

Step-by-step explanation:

For the first equation, we have to assume that you will spend all of your money, otherwise it becomes an inequation. The money you spend on regular bars is 0.75x dollars and the money you spend on strawberry bars is 1.25y, so if you spend your 3.75 dollars on the bars, then 0.75x+1.25y=3.75.

For the second, you will always buy x+y health bars, regular and strawberry. There isn't enough information to make this into a equation, the only thing we can deduce is the inequation x+y>=3.

If we also assume that x and y are integers (we can't buy half-bars or one-fourth of a bar) then the minimum number of bars we can buy is 3 (3 strawberry bars) and the maximum is 5 bars (5 regular bars). x+y must be an integer too, so the possibilities for the second equation are x+y=3, x+y=4 and x+y=5. There is a finite number of solutions in any case.  

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Answer:

The volume of the sphere is 1436 in³

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Step-by-step explanation:

radius = half of diameter

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r = 14ft / 2 = 7ft

To calculate the volume of a sphere we have to use the following formula:

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π = 3.14

V = ⁴⁄₃πr³

we replace with the known values

V = ⁴⁄₃ * 3.14 * (7ft)³

V = 4.187 * 343 in³

V = 1436 in³

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|3 + y| - 4 = 7 solve
vfiekz [6]

Answer:

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Step-by-step explanation:

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alexgriva [62]

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f(4)+f(5)+f(6)+f(7)=0.35

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E[X]=\displaystyle\sum_xx\,f(x)=3.16

Since X is the number of absent students on Monday, the expectation E[X] is the number of students you can expect to be absent on average on any given Monday. According to the distribution, you can expect around 3 students to be consistently absent.

c) The variance is

V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2

where

E[X^2]=\displaystyle\sum_xx^2\,f(x)=11.58

So the variance is

V[X]=11.58-3.16^2\approx1.59

The standard deviation is the square root of the variance:

\sqrt{V[X]}\approx1.26

d) Since Y=7X+3 is a linear combination of X, computing the expectation and variance of Y is easy:

E[Y]=E[7X+3]=7E[X]+3=25.12

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e) The covariance of X and Y is

\mathrm{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]

We have

XY=X(7X+3)=7X^2+3X

so

E[XY]=E[7X^2+3X]=7E[X^2]+3E[X]=90.54

Then the covariance is

\mathrm{Cov}[X,Y]=90.54-3.16\cdot25.12\approx11.16

f) Dividing the covariance by the variance of X gives

\dfrac{\mathrm{Cov}[X,Y]}{V[X]}\approx\dfrac{11.16}{1.59}\approx0.9638

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