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3 years ago

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A random sample of CO2 levels in a school has a sample mean of x¯=598.4 ppm and sample standard deviation of s=86.7 ppm. Use the

Empirical Rule to determine the approximate percentage of CO2 levels that lie between 338.3 and 858.5 ppm. Round your answer to the nearest tenth.

1 answer:

astraxan [27]3 years ago

6 0

Answer:

99.7%

Step-by-step explanation:

we start by getting the z-scores for the values given

We can get the z-scores using the formula below;

z-scores = (x-mean)/SD

for 338.5;

z = (338.5-598.4)/86.7 = -2.99 approximately-3

for 858.5;

z = (858.5-598.4)/86.7 = 3

so we want to get the percentages that lie within 3 standard deviations from the mean

According to the empirical rule, the percentages within 3 SD of the mean is;

100-0.15-0.15 = 99.7%

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<u><em></em></u> $sin(A) = \frac{opposite side}{Hypotenuse} = \frac{a}{c} = \frac{4}{5}$ <u><em></em></u>

<u><em></em></u> $cos(A) = \frac{Adjacent side}{Hypotenuse} = \frac{b}{c} = \frac{3}{5}$ <u><em></em></u>

<u><em></em></u> $tan(A) = \frac{opposite side}{Adjacent side} = \frac{a}{b} = \frac{4}{3}$ <u><em></em></u>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given b = 3 and hypotenuse c = 5

Given ΔABC is a right angle triangle

By using pythagoras theorem

c² = a² + b²

⇒ a² = c² - b²

⇒ a² = 5²-3²

=25 - 9

a² = 16

⇒ a = √16 = 4

The sides of right angle triangle a = 4 ,b = 3 and c = 5

<u><em>Step(ii):-</em></u>

<u><em></em></u> $sin(A) = \frac{opposite side}{Hypotenuse} = \frac{a}{c} = \frac{4}{5}$ <u><em></em></u>

<u><em></em></u> $cos(A) = \frac{Adjacent side}{Hypotenuse} = \frac{b}{c} = \frac{3}{5}$ <u><em></em></u>

<u><em></em></u> $tan(A) = \frac{opposite side}{Adjacent side} = \frac{a}{b} = \frac{4}{3}$ <u><em></em></u>

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