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Gnoma [55]
2 years ago
9

Tre has to pay $180 to book the recording studio plus $30 per hour of recording time. The function c(h)=30h+180 relates the tota

l cost, c(h) of the recording session (in dollars) to the number of hours, (h) needed to record the tracks.
Mathematics
1 answer:
blagie [28]2 years ago
8 0

Answer:

  • 1. $330
  • 2. 11.5 hours

Step-by-step explanation:

<u><em>Continuation of the question:</em></u>

  • <em>1. Find c (5) and interpret your solution in the context of the problem </em>
  • <em>2. Find the value of h when c() = 525 and interpret your solution in the context of the problem</em>

<u>Solution</u>

  • c(h) = 30h + 180

1 ...............................................

  • c(5) = 30*5 + 180 = $330

2 ...............................................

  • 525 = 30h + 180
  • 30h = 525 - 180
  • 30h = 345
  • h = 345/30
  • h = 11.5 hours

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3 years ago
Let set A = {odd numbers between 0 and 100} and set B = {numbers between 50 and 150 that are evenly divisible by 5}. What is A ∩
fgiga [73]
<span>A = {odd numbers between 0 and 100}
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The notation </span><span>A ∩ B means the set of items that are in set A and also in set B. In terms of venn diagrams, it's the overlapping region between circle A and circle B

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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
Katyanochek1 [597]

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

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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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