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NARA [144]
3 years ago
9

What is the weight (in grams) of a liquid that exactly fills a 182.8 milliliter container if the density of the liquid is 0.135?

Round to the nearest hundredth when necessary, and only enter numerical values, which can include a decimal point.
Mathematics
2 answers:
Setler79 [48]3 years ago
7 0

Answer:

27.68 gram

Step-by-step explanation:

Now, 182.8 milliliters of the container is completely filled up by the same liquid, and we have to find the weight of 182.8 milliliters of liquid.

If the weight of 1 milliliter of the liquid is 0.135 grams.

Then the weight of 182.8 milliliters of liquid will be (0.135 × 182.8) = 24.678 ≈ 27.68 grams (Rounded to the nearest hundredth) (Answer)

Click to let others know, how helpful is it

AleksAgata [21]3 years ago
6 0

Answer:

24.68

Step-by-step explanation:

Multiply 182.8 by 0.135 and then round.

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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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