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Flura [38]
4 years ago
14

Kwamee created a coffee blend for his cafe by mixing Kona beans and Fuji beans.Kona beans cost $11 per pound and Fuji beans cost

$7.50 per pound.He bought a total of 23 pounds of coffee beans and it cost $197. Write a system of equations that can be used to determine how many pounds of Kona and Fuji beans Kwamee bought.
Mathematics
1 answer:
Svetlanka [38]4 years ago
8 0
Let's use K for Kona and F for Fuji.  The system of equations has to be a balanced system.  For example, you can't mix the number of pounds of beans with the cost for each because pounds and dollars are different and you can only combine like terms...pounds with pounds and dollars with dollars.  So let's start with the number of pounds.  Since we don't know how much of each he bought we have the 2 unknowns, F and K, but we DO know that he bought 23 pounds total. So the first equation is
K + F = 23
Now let's see what we can do with the dollars. Again, we don't know how much he bought of each kind of coffee, but we do know that Kona beans cost $11 per pound and that Fuji beans cost $7.50 per pound, and we know that he spent a total of $197. So let's set that up:
11K + 7.50F = 197
Those are your 2 equations. It doesn't say you need to solve them, so you're done.

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3 years ago
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genetic experiment with peas resulted in one sample of offspring that consisted of green peas and yellow peas. a. Construct a ​%
Andreas93 [3]

Complete Question

A genetic experiment with peas resulted in one sample of offspring that consisted of 432 green peas and 164 yellow peas. a. Construct a 95% confidence interval to estimate of the percentage of yellow peas. b. It was expected that 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations?

Answer:

The  95%  confidence interval is  0.2392  <  p < 0.3108

No, the confidence interval includes​ 0.25, so the true percentage could easily equal​ 25%

Step-by-step explanation:

From the question we are told that

  The total sample size is  n  =  432 + 164 =596

   The  number of  offspring that is yellow peas is y =  432

   The  number of  offspring that is green peas   is g =  164

   

The sample proportion for offspring that are yellow peas is mathematically evaluated as

        \r p  =  \frac{ 164 }{596}

        \r p  =  0.275

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

       \alpha  =  (100 - 95)\%

      \alpha =  5\%  =  0.0 5

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

      Z_{\frac{\alpha }{2} } = 1.96

Generally the margin of error is mathematically evaluated as

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

=>      E = 1.96 * \sqrt{\frac{0.275 (1- 0.275 )}{596} }

=>      E =  0.0358

The  95%  confidence interval is mathematically represented as

      \r p - E  <  p < \r p + E

=>   0.275 -  0.0358  <  p < 0.275 +  0.0358

=>   0.2392  <  p < 0.3108

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