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PolarNik [594]
3 years ago
15

Mary has twice as many dogs as David has cats and josh has enough animal collars to give a collar to one third of the dogs or to

give a collar to all the cats except one. How many collars does josh have?
Mathematics
1 answer:
Ann [662]3 years ago
3 0
D = 2C
J = 1/3D
J = C - 1

J = 1/3D
J = 1/3(2C)
J = 2/3C

2/3C = C - 1
2/3C - 3/3C = -1
-1/3C = -1
C = -1 * -3
C = 3...there are 3 cats

J = C - 1
J = 3 - 1
J = 2.....there are 2 collars





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4 years ago
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Find f (2b^2) for (x)=x^2-4x
BlackZzzverrR [31]

Answer:

A

Step-by-step explanation:

We are given the function:

f(x) = x^2 - 4x

And we want to find:

\displaystyle f(2b^2)

Substitute:

\displaystyle f(2b^2) =(2b^2)^2 -4(2b^2)

And evaluate:

\displaystyle \begin{aligned}  f(2b^2) &=(2b^2)^2 -4(2b^2) \\ &=(4b^4) +(-8b^2) \\ &= 4b^4 - 8b^2 \end{aligned}

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7 0
3 years ago
A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex
Olenka [21]

Answer:

a

 The  90% confidence interval that  estimate the true proportion of students who receive financial aid is

     0.533  <  p <  0.64

b

   n = 1789

Step-by-step explanation:

Considering question a

From the question we are told that

      The sample size is  n = 200

      The number of student that receives financial aid is k = 118

Generally the sample proportion is  

      \^ p = \frac{k}{n}

=>   \^ p = \frac{118}{200}

=>   \^ p = 0.59

From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

Generally from the normal distribution table the critical value  of \frac{\alpha }{2}  is  

   Z_{\frac{\alpha }{2} } =  1.645

Generally the margin of error is mathematically represented as  

     E =  Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }

 =>E =  1.645 * \sqrt{\frac{0.59 (1- 0.59)}{200} }

=>  E = 0.057

Generally 90% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

  =>  0.533  <  p <  0.64  

Considering question b

From the question we are told that

    The margin of error  is  E = 0.03

From the question we are told the confidence level is  99% , hence the level of significance is    

      \alpha = (100 - 99 ) \%

=>   \alpha = 0.01

Generally from the normal distribution table the critical value  of   is  

   Z_{\frac{\alpha }{2} } = 2.58

Generally the sample size is mathematically represented as      

        [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )

=>      n = [\frac{2.58}{0.03} ]^2 * 0.59 (1 - 0.59 )

=>      n = 1789

8 0
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AnnyKZ [126]

Answer:

$152.28

Step-by-step explanation:

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94 x 62% = 94 x .62 = 58.28

In order to determine the sale price, add the markup to the original price.

94 + 58.28 = $152.28

3 0
3 years ago
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