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Advocard [28]
2 years ago
14

You sell sporting goods. Your wages depend on the value of your sales. One week you sold $3,500 in sporting goods, earning $950.

Another week you sold $2,800 in sporting goods, earning $810. Write an equation of the line that represents the situation in standard form.
Mathematics
1 answer:
Semmy [17]2 years ago
7 0

Answer:

y = 0.2x + 250

Step-by-step explanation:

let the sales be x and y be earnings

thus,

given

x₁ = $3,500 ; y₁ = $950

and,

x₂ = $2,800 ; y₂ = $810

Now,

the standard line equation is given as:

y = mx + c

here,

m is the slope

c is the constant

also,

m = \frac{y_2-y_1}{x_2-x_1}

or

m = \frac{810-950}{\textup{2,800-3,500}}

or

m = 0.2

substituting the value of 'm' in the equation, we get

y = 0.2x + c

now,

substituting the x₁ = $3,500 and y₁ = $950 in the above equation, we get

$950 = 0.2 × $3,500 + c

or

$950 = $700 + c

or

c = $250

hence,

The equation comes out as:

y = 0.2x + 250

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Step-by-step explanation:

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2 years ago
Kellan collected two sets of data and recorded them in the table. 
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Answer:

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Explanation:

We are given the below two sets of data:

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Now, let's find the range of Set 1:

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2 years ago
Read 2 more answers
. Given f(x) = e 2x e 2x + 3e x + 2 : (a) Make the substitution u = e x to convert Z f(x) dx into an integral in u (HINT: The ea
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Answer:

Step-by-step explanation:

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f(x)=\frac{e^{2x}}{e^{2x}+3e^x+2}

a)

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b)

Apply partial fraction in (a), we get;

\frac{u}{u^2+3u+2}=\frac{2}{u+2}-\frac{1}{u+1}\\\\\therefore u^2+3u+2\\=u^2+2u+u+2\\=u(u+2)+1(u+1)\\=(u+2)(u+1)\\\\Now\,\int\frac{u}{u^2+3u+2}\,du=\int\frac{2}{u+2}du-\int\frac{1}{u+1}du\\\\=2ln|u+2|-ln|u+1|+c\\=2ln|e^x+2|-lm|e^x+1|+c

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2 years ago
Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your methods as well
laiz [17]

Answer:

The  value is  P(A) =  0.133617

Step-by-step explanation:

From the question we are told that

  The  mean is  \mu =  7.5

  The standard deviation is  \sigma  =  0.2

  The safest water level is  between  7.2 and  7.8

Generally the probability that the selected pool has a pH level that is not considered safe is mathematically represented as

       P(A) =  1 - P(7.2 \le X  \le 7.8 )

Here  

      P(7.2 < X  < 7.8 ) = P(\frac{ 7.2 - \mu }{\sigma } <  \frac{X - \mu }{ \sigma }

Generally \frac{X - \mu }{ \sigma } =  Z (The  \ standardized \  value  \  of  X )

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=>  P(7.2 < X  < 7.8 ) = P(Z <  1.5) - P(  Z < - 1.5)

From the z-table the probability of  (Z <  -1.5) and   (  Z <1.5)  are

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So

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=>  P(A) =  0.133617

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