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Yuri [45]
3 years ago
7

Jose wants to be a vet, so he reads a lot. This month he read 14 books about dogs, 5 books about birds,6 about other animals and

7 biographies. Of the books he read, how many more were about animals than people?
Mathematics
1 answer:
Lana71 [14]3 years ago
7 0
18 more were about animals
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What are the common factors of 60 and 72?
ddd [48]
Answer: 1, 2, 3, 4, 6, 12

Calculation/Explanation:

Common factors are factors that both the numbers, in this case, 60 and 72, have in common. The factors listed above are all factors shared by 60 and 72.
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What's the value of x <br><br><br><br> (Middle School)
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The answer is 360 hope it helps
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The scale factor between two similar shapes is 4. what is the scale factor of area
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The amount of money spent on textbooks per year for students is approximately normal.
Ostrovityanka [42]

Answer:a

a

   336.04    <  \mu < 443.96

b

  The  margin of error will increase

c

The  margin of error will decreases

d

The 99% confidence interval is  0.4107 <  p  < 0.4293

Step-by-step explanation:

From the question we are  told that

   The sample size  n =  19

    The sample mean is  \= x  = \$\  390

    The  standard deviation is  \sigma =  \$ \  120

 

Given that the confidence level is  95% then the level of significance is mathematically represented as

           \alpha = 100 -  95

          \alpha  =  5 \%

          \alpha  =  0.05

Next we obtain the critical value of \frac{\alpha }{2} from the normal distribution table

    So  

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

The  margin of error is mathematically represented as

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

=>    E = 1.96 *  \frac{120}{\sqrt{19} }

=>   E = 53.96

The 95% confidence interval is  

     \= x  -  E  <  \mu < \= x  +  E

=>   390  -   53.96   <  \mu < 390  -   53.96

=>  336.04    <  \mu < 443.96

When the confidence level increases the Z_{\frac{\alpha }{2} } also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

         n  \  \ \alpha  \ \  \frac{1}{E^2 }

So if the sample size increases the margin of error decrease

The  sample proportion is mathematically represented as

       \r p  =  \frac{210}{500}

       \r p  = 0.42

Given that the confidence level is 0.99 the level of significance is  \alpha =  0.01

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

      Z_{\frac{\alpha }{2} }  =  2.58

  Generally the margin of error is mathematically represented as

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

=>   E =  0.42 *  \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }

=>     E =  0.0093

The 99% confidence interval  is

     \r p  -  E <  p  < \r p  +  E

     0.42  -  0.0093 <  p  < 0.42  +  0.0093

     0.4107 <  p  < 0.4293

 

4 0
3 years ago
Word problem
Vlad [161]

Answer:

p=2

Step-by-step explanation:

4.05p+14.40=4.50(p+3)          < equation

4.05p+14.40=4.50p+13.50     < multiply

14.40=.45p+13.50                   < subtract

.9=.45p                                    < subtract

2=p                                          < divide

5 0
3 years ago
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