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musickatia [10]
3 years ago
14

Jon makes a mix of salad that cost $15 using as much apples that cost $4/kg as oranges that cost $6/kg. How much of each fruit i

s in the salad?
Mathematics
2 answers:
alexira [117]3 years ago
7 0
<span>Jon used 1.5 kg of apples and 1.5 kg of oranges. If 1 kg of apples cost $4, then 1.5 kg will cost $6. If 1 kg of oranges costs $6, then 1.5 kg of oranges will cost $9. $6 + $ 9 is equal to $15 which is the total Jon spent.</span>
marysya [2.9K]3 years ago
6 0
Let
 x = number of apples.
 y = number of oranges.
 we have to write the following equation to represent the problem:
 4x + 6y = 15
 To satisfy the equation, Jon must have used
 x = 2.25
 y = 1
 Substituting
 4 (2.25) +6 (1) = 15
 9 + 6 = 15
 answer
 Jhon used 2.25kg of apple and 1kg of orange to make the salad.
 Note: Since the problem does not have any other restrictions, there may be several apple and orange combinations that cost $ 15 per salad.
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Answer:

1) Null hypothesis:p\geq 0.41  

Alternative hypothesis:p < 0.41

2) \hat p=0.36 estimated proportion of people indicated that they watch the late evening news on this local CBS station

3) z_{crit}=-2.33

And we can use the following excel code to find it: "=NORM.INV(0.01,0,1)"

4) z=\frac{0.36 -0.41}{\sqrt{\frac{0.41(1-0.41)}{1000}}}=-1.017  

5) z_{crit}=-1.28

And we can use the following excel code to find it: "=NORM.INV(0.1,0,1)"

6) We see that |t_{calculated}| so then we have enough evidence to FAIL to reject the null hypothesis at 1% of significance.

7) Null hypothesis:p\geq 0.41  

Step-by-step explanation:

Data given and notation  

n=100 represent the random sample taken

X represent the people indicated that they watch the late evening news on this local CBS station

\hat p=0.36 estimated proportion of people indicated that they watch the late evening news on this local CBS station

p_o=0.41 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v{/tex} represent the p value (variable of interest)  Part 1We need to conduct a hypothesis in order to test the claim that 11:00 PM newscast reaches 41 % of the viewing audience in the area:  Null hypothesis:[tex]p\geq 0.41  

Alternative hypothesis:p < 0.41

Part 2  

\hat p=0.36 estimated proportion of people indicated that they watch the late evening news on this local CBS station

Part 3

Since we have a left tailed test we need to see in the normal standard distribution a value that accumulates 0.01 of the area on the left and on this case this value is :

z_{crit}=-2.33

And we can use the following excel code to find it: "=NORM.INV(0.01,0,1)"

Part 4

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.36 -0.41}{\sqrt{\frac{0.41(1-0.41)}{1000}}}=-1.017  

Part 5

Since we have a left tailed test we need to see in the normal standard distribution a value that accumulates 0.1 of the area on the left and on this case this value is :

z_{crit}=-1.28

And we can use the following excel code to find it: "=NORM.INV(0.1,0,1)"

Part 6

We see that |t_{calculated}| so then we have enough evidence to FAIL to reject the null hypothesis at 1% of significance.

Part 7

Null hypothesis:p\geq 0.41  

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