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swat32
4 years ago
14

X-2y=4 x-2y=5 how many solutions does these equations have

Mathematics
1 answer:
cupoosta [38]4 years ago
6 0

Answer:

No solutions

Step-by-step explanation:

Rewrite x-2y=4 x-2y=5 in column form:

x-2y=4

x-2y=5

Because both equations have x - 2y on the left and different constants on the right, we know immediately that there is no solution.  These two lines are parallel and different; they never cross.

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For each level of precision, find the required sample size to estimate the mean starting salary for a new CPA with 95 percent co
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Answer:

(a) Margin of error ( E) = $2,000 , n = 54

(b)   Margin of error ( E) = $1,000 , n = 216

(c)   Margin of error ( E) = $500 , n= 864

Step-by-step explanation:

Given -

Standard deviation \sigma = $7,500

\alpha = 1 - confidence interval = 1 - .95 = .05

Z_{\frac{\alpha}{2}} =  Z_{\frac{.05}{2}} = 1.96

let sample size is n

(a) Margin of error ( E) = $2,000

Margin of error ( E)  = Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}

                           E   = Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}

Squaring both side

E^{2} = 1.96^{2}\times\frac{7500^{2}}{n}

n =\frac{1.96^{2}}{2000^{2}} \times 7500^{2}

n =  54.0225

n = 54 ( approximately)

(b)   Margin of error ( E) = $1,000

          E     = Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}

         1000   =  Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}

Squaring both side

1000^{2} = 1.96^{2}\times\frac{7500^{2}}{n}

n =\frac{1.96^{2}}{1000^{2}} \times 7500^{2}

n = 216

(c)   Margin of error ( E) = $500

   E = Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}

  500 = Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}

Squaring both side

500^{2} = 1.96^{2}\times\frac{7500^{2}}{n}

n =\frac{1.96^{2}}{500^{2}} \times 7500^{2}

n = 864

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4 years ago
At the end of the baseball season, four families decided to buy the coach a gift certificate. Each family contributed $35 toward
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Step-by-step explanation:

3 0
3 years ago
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maks197457 [2]

Answer:

$7,305.50

Step-by-step explanation:

The computation of the FICA tax pay in 2005 is shown below:

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the area of a triangle is 124 square units. what would it's new area be if its base was half as long and its height was three ti
Montano1993 [528]
To solve this problem you must apply the proccedure shown below:

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 bh/2=124
 bh=124x2
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 2. The problem asks for the new area of the triangle <span>if its base was half as long and its height was three times as long. Then, you have:

 Base=b/2
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 A'=(b/2)(3)/2
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 4. When you substitute bh=248 into </span>A'=3bh/4, you obtain:
<span>
 A'=186 units</span>²
<span>
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4 years ago
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