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Anna11 [10]
2 years ago
13

Lixin intends to buy either Gift A, which costs $10 or Gift B, which costs $8, as Christmas gifts for each of her parents, 2 sib

lings, 13 relatives and 10 friends. Given that she intends to spend $230, find the number of each gift she should buy
Please help !! I need someone
Mathematics
1 answer:
sasho [114]2 years ago
3 0
10a + 8b = 230
a + b = (2 + 2 + 13 + 10)...a + b = 27 (this is the number of gifts to buy)

a = 27 - b

10(27 - b) + 8b = 230
270 - 10b + 8b = 230
-10b + 8b = 230 - 270
-2b = -40
b = -40/-2
b = 20 <=== she bought 20 eight dollar gifts

a = 27 - b
a = 27 - 20
a = 7 <=== she bought 7 ten dollar gifts
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Depth of the bathtub = 18 inches

Time taken to fill 3 inches of the bathtub = 2 minutes

Then

The amount of depth needed to fill with water = (18-3) inches

                                                                        = 15 inches

Time required to fill 15 inches of the bathtub = (15*2)/3 minutes

                                                                      = 30/3 minutes

                                                                      = 10 minutes

So John is correct in thinking that it will take 10 more minutes for the water to reach the top of the tub if he continues at the same rate.

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I hope this helps you



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Hard times In June 2010, a random poll of 800 working men found that 9% had taken on a second job to help pay the bills. (www.ca
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Answer:

a) The 95% confidence interval would be given (0.0702;0.1098).

We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)

b) Since the confidence interval NOT contains the value 0.06 we  have anough evidence to reject the claim at 5% of significance.  

Step-by-step explanation:

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".  

p represent the real population proportion of interest  

\hat p =0.09 represent the estimated proportion for the sample  

n=800 is the sample size required  

z represent the critical value for the margin of error  

Confidence =0.95 or 95%

Part a

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

The confidence interval would be given by this formula  

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.96  

The margin of error is given by :

Me=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

Me=1.96 \sqrt{\frac{0.09(1-0.09)}{800}}=0.0198

And replacing into the confidence interval formula we got:  

0.09 - 1.96 \sqrt{\frac{0.09(1-0.09)}{800}}=0.0702  

0.09 + 1.96 \sqrt{\frac{0.108(1-0.09)}{800}}=0.1098  

And the 95% confidence interval would be given (0.0702;0.1098).

We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)

Part b

Since the confidence interval NOT contains the value 0.06 we  have anough evidence to reject the claim at 5% of significance.

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