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mariarad [96]
2 years ago
13

Please answer it’s due tomorrow

Mathematics
2 answers:
zysi [14]2 years ago
5 0

Step-by-step explanation:

First method :-

<u>Using unitary method</u><u>,</u>

5/2 scoops = 10 servings

1 scoop = 10/5/2 = 4 servings

15 scoops = 15 * 4 = 60 servings

15 scoops = 60 servings of bannock

Second method :-

<u>Using proportions,</u>

Let x be servings of bannock for 15 scoops

Let ratios be 5/2 : 10 and 15 : x

Now,

5/2 : 10 = 15 : x

or, 5/2/10 = 15/x

or, 5/2 * 1/10 = 15/x

or, 1/4 = 15/x

or, x = 15 * 4

so, x = 60

So,

15 scoops = 60 servings of bannock

malfutka [58]2 years ago
4 0

Answer:

60 servings

Step-by-step explanation:

Method 1: 2.5/10 (fraction) = 15/x (fraction)

Draw an arrow from 2.5 to 15 and write x6 (times 6)

Do the same from 10 to x

Then do 10*6 = 60

I don't know another method but hope this helps!

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From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean leng
alexandr402 [8]

Complete Question

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean length of bass caught in a small lake. A normal distribution was assumed. Using the 90% confidence interval obtain:

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b. A 95% confidence interval for \mu.

Answer:

a

\= x  = 13.4   .   E = 2.3

b

10.7 <  \mu < 16.1

Step-by-step explanation:

From the question we are told that

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  The 90% confidence interval is  (11.1, 15.7)

Generally the point estimate of  \mu is mathematically  evaluated  as

       \= x  = \frac{11.1 + 15.7 }{2}

=>    \= x  = 13.4

Generally the margin of error is mathematically evaluated  as

     E = \frac{15.7 - 11.1}{2 }

=> E = 2.3

  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 equation for the lower limit of the confidence interval is  

      \= x  -  Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{18} } = 11.1

=> 13.4   -  0.3877 s  = 11.1

=>  s = 5.932

  From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

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

   Z_{\frac{\alpha }{2} } =  1.96

Generally the margin of error is mathematically represented as  

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

=>    E =  1.96 *  \frac{5.932}{\sqrt{18} }

=>    E =  2.7      

Generally 95% confidence interval is mathematically represented as  

      \= x -E <  \mu <  \=x  +E

=>    13.4  -  2.7  <  \mu < 13.4  +   2.7

=>    10.7 <  \mu < 16.1

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