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4 years ago

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An isosceles triangle has an angle that measures 68°. Which could be the measure of another angle in the triangle? A. 22° B. 56°

C. 60° D. 112°

1 answer:

Lisa [10]4 years ago

5 0

Answer:

B

Step-by-step explanation: 68+x+x=180. subtract 68 from both sides, and simplify the x's. so 2x=112. than divide by 2. x=56

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There are 12 grams of sugar in 1⁄3 of a piece of candy. How much sugar is in 3⁄4 of a piece of candy?

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3 years ago

The amounts (in ounces) of randomly selected eight 16-ounce beverage cans are given below. See Attached Excel for Data. Assume t

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The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

The amounts (in ounces) of randomly selected eight 16-ounce beverage cans are given below.

16.5, 15.2, 15.4, 15.1, 15.3, 15.4, 16, 15.1

Assume that the amount of beverage in a randomly selected 16-ounce beverage can has a normal distribution. Compute a 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans and fill in the blanks appropriately.

A 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is ( , ) ounces. (round to 3 decimal places)

Answer:

$99\% \: \text {confidence interval} = (14.886, \: 16.113)\\\\$

Therefore, the 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (14.886, 16.113) ounces.

Step-by-step explanation:

Let us find out the mean amount of the 16-ounce beverage cans from the given data.

Using Excel,

=AVERAGE(number1, number2,....)

The mean is found to be

$\bar{x} = 15.5$

Let us find out the standard deviation of the 16-ounce beverage cans from the given data.

Using Excel,

=STDEV(number1, number2,....)

The standard deviation is found to be

$s = 0.4957$

The confidence interval is given by

$\text {confidence interval} = \bar{x} \pm MoE\\\\$

Where $\bar{x}$ is the sample mean and Margin of error is given by

$$ MoE = t_{\alpha/2} \cdot (\frac{s}{\sqrt{n} } ) $ \\\\$

Where n is the sample size, s is the sample standard deviation and is the t-score corresponding to a 99% confidence level.

The t-score corresponding to a 99% confidence level is

Significance level = α = 1 - 0.99 = 0.01/2 = 0.005

Degree of freedom = n - 1 = 8 - 1 = 7

From the t-table at α = 0.005 and DoF = 7

t-score = 3.4994

$MoE = t_{\alpha/2}\cdot (\frac{s}{\sqrt{n} } ) \\\\MoE = 3.4994 \cdot \frac{0.4957}{\sqrt{8} } \\\\MoE = 3.4994\cdot 0.1753\\\\MoE = 0.6134\\\\$

So the required 99% confidence interval is

$\text {confidence interval} = \bar{x} \pm MoE\\\\\text {confidence interval} = 15.5 \pm 0.6134\\\\\text {confidence interval} = 15.5 - 0.6134, \: 15.5 + 0.6134\\\\\text {confidence interval} = (14.886, \: 16.113)\\\\$

Therefore, the 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (14.886, 16.113) ounces.

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4 years ago

In triangle JKL, tan(b°) = 3/4 and cos(b°) =4/5. If triangle JKL is dilated by a scale factor of 1/2, what is sin(b°)?

zysi [14]

Answer:

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Step-by-step explanation:

It is given that,

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We know that,

$\dfrac{\sin \theta}{\cos \theta}=\tan \theta$

$\dfrac{\sin (b^\circ)}{\cos (b^\circ)}=\tan (b^\circ)$

$\sin (b^\circ)=\tan (b^\circ)\times \cos (b^\circ)$

$\sin (b^\circ)=\dfrac{3}{4}\times \dfrac{4}{5}$

$\sin (b^\circ)=\dfrac{3}{5}$

Therefore, $\sin (b^\circ)=\dfrac{3}{5}$ .

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