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goldenfox [79]
3 years ago
15

Every evening jenna empties her pockets and puts the change in a jar at the end of the week she had 38 coins all of them dimes a

nd quarters when she added them up she had a totals of $6.95
Mathematics
2 answers:
ser-zykov [4K]3 years ago
8 0
I imagine the part you left out was "how many of each coin did she have?".  I'm going with that.  The thing is you have to relate the NUMBER of coins to each other, which is very different from the VALUE of the coins.  I could have 1 single quarter, but it's worth 25 cents.  Number vs. value.  If she had 38 coins made up of dimes and quarters, then d + q = 38.  That's the number of dimes and quarters, both still unknown.  A dime is worth .10 and a quarter is .25.  Therefore, .10d + .25q = 6.95.  That's the value of the coins.  Solve the first equation for dimes.  d = 38 - q.  Now we can sub in that value of d into the second equation, so we have only one unknown instead of 2.  .10(38-q)+.25q=6.95.  Distribute through the parenthesis to get 3.8-.1q+.25q=6.95.  Combine all the like terms on both sides to get.15q=3.15.  Solve for q and get that q = 21.  So we have 21 quarters.  38-21 = # of dimes, which is 17.  There you go!  Hope that's what you actually needed!  ;)
dedylja [7]3 years ago
7 0
A) d + q = 38
B)  .1d +.25q = 6.95
Multiply A) by -.1
A) -.1d + -.1q = -3.8The adding B)B)  .1d +.25q = 6.95
.15q = 3.15
21 quarters17 dime
Double-Check********************************************************
17 * .10 + 21 * .25 = 
1.70 + 5.25 = 
6.95

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The mean cost of a meal for two in a mid-range restaurant in Tokyo is $40 (Numbeo.com website, December 14, 2014). How do prices
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Answer:

a) Me=2.02 \frac{6.827}{\sqrt{42}}=2.13

b) So on this case the 95% confidence interval would be given by (30.53;34.79)    

c) On this case the confidence interval not contains the price $40, so we can conclude that the prices for Hong Kong mid-range restaurants are significant less than the prices for mid range restaurants in Tokyo at 5 % of significance.

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

2) Part a

The margin of error is given by:

Me=t_{\alpha/2}\frac{s}{\sqrt{n}}

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=42-1=41

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,41)".And we see that t_{\alpha/2}=2.02

Me=2.02 \frac{6.827}{\sqrt{42}}=2.13

3) Part b

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=32.66

The sample deviation calculated s=6.827

Now we have everything in order to replace into formula (1):

32.66-2.02\frac{6.827}{\sqrt{42}}=30.532    

32.66+2.02\frac{6.827}{\sqrt{42}}=34.788

So on this case the 95% confidence interval would be given by (30.532;34.788)    

4) Part d

On this case the confidence interval not contains the price $40, so we can conclude that the prices for Hong Kong mid-range restaurants are significant less than the prices in mid range restaurants in Tokyo at 5 % of significance.

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