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Mashcka [7]
3 years ago
15

I have 2 questions. 1. Find the mode and the median for this data set. 5,7,5,8,9,5,8. 2. Write 543.24 in expanded form.

Mathematics
1 answer:
katen-ka-za [31]3 years ago
3 0
To find the median, put the numbers in order to least to greatest, which would be 5, 5, 5, 7, 8, 8, 9. the median is the number in the middle, which is seven. the mode is the number that occurs most often in the set, which is 5 in this case (appears three times). 2. 543.24 in expanded form is essentially pulling the number apart into its different values places. in this case, it would be five hundred plus forty plus three plus two tenths plus four hundredths, which is 500 + 40 + 3 + .2 + .4. hope this helps!
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Whats -5/6 divided by 3/10
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3 years ago
Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorab
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Answer:

(a) Customer will not purchase the light bulbs at significance level of 0.05

(b) Customer will purchase the light bulbs at significance level of 0.01 .

Step-by-step explanation:

We are given that Light bulbs of a certain type are advertised as having an average lifetime of 750 hours. A random sample of 50 bulbs was selected,  and the following information obtained:

Average lifetime = 738.44 hours and a standard deviation of lifetimes = 38.2 hours.

Let Null hypothesis, H_0 : \mu = 750 {means that the true average lifetime is same as what is advertised}

Alternate Hypothesis, H_1 : \mu < 750 {means that the true average lifetime is smaller than what is advertised}

Now, the test statistics is given by;

       T.S. = \frac{Xbar-\mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, X bar = sample mean = 738.44 hours

               s  = sample standard deviation = 38.2 hours

               n = sample size = 50

So, test statistics = \frac{738.44-750}{\frac{38.2}{\sqrt{50} } } ~ t_4_9

                            = -2.14

(a) Now, at 5% significance level, t table gives critical value of -1.6768 at 49 degree of freedom. Since our test statistics is less than the critical value of t, so which means our test statistics will lie in the rejection region and we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is smaller than what is advertised and so consumer will not purchase the light bulbs.

(b) Now, at 1% significance level, t table gives critical value of -2.405 at 49 degree of freedom. Since our test statistics is higher than the critical value of t, so which means our test statistics will not lie in the rejection region and we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is same as what it has been advertised and so consumer will purchase the light bulbs.

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Plz help me I've been stuck on this
timurjin [86]
I found this :) hope it helps

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