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Andre45 [30]
3 years ago
6

Mhanifa can you please help? This is due soon. Look at the picture attached. I will mark brainliest!

Mathematics
1 answer:
TEA [102]3 years ago
5 0

Answer:

  • 84°
  • 125°

Step-by-step explanation:

<u>Sum of the interior angles of a regular polygon:</u>

  • S(n) = 180°(n - 2), where n- number of sides
<h3>Exercise 4</h3>

<u>Pentagon has sum of angles:</u>

  • S(5) = 180°(5 - 2) = 540°

<u>Sum the given angles and find x:</u>

  • x° + 122° + 100° + 90° + 144° = 540°
  • x° + 456° = 540°
  • x° = 540° - 456°
  • x° = 84°
<h3>Exercise 5</h3>

<u>Hexagon has sum of angles:</u>

  • S(6) = 180°(6 - 2) = 720°

<u>Sum the given angles and find x:</u>

  • x° + 110° + 160° + 105° + 105° + 115° = 720°
  • x° + 595° = 720°
  • x° = 720° - 595°
  • x° = 125°
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The answer is $1,026.

Solving: Use Simple interest formula, I=Prt.

Make the rate into a decimal (1.6% -> 0.016)

plug in the numbers into the formula. 950 x 0.016 x 5 = 76$.

76$ will be our interest, now to find the balance we would have to add the the starting money (950) with our interest (76$), resulting in our answer of $1,026 after 5 years.
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Step-by-step explanation:

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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:

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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

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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.

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