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2 years ago

In the figure, mAB = 45° and mCD = 23°. The diagram is not drawn to scale.

Mathematics

1 answer:

Yakvenalex [24]2 years ago

4 0

Answer:

Option A. $x=34\°$

Step-by-step explanation:

we know that

The measure of the inner angle is the semi-sum of the arcs comprising it and its opposite.

$x=\frac{1}{2}(arc\ CD+arc\ AB)$

substitute the values

$x=\frac{1}{2}(23\°+45\°)$

$x=34\°$

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The time for a professor to grade a student's homework in statistics is normally distributed with a mean of 12.6 minutes and a s

Rus_ich [418]

Answer:

37.23% probability that randomly selected homework will require between 8 and 12 minutes to grade

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 12.6, \sigma = 2.5$

What is the probability that randomly selected homework will require between 8 and 12 minutes to grade?

This is the pvalue of Z when X = 12 subtracted by the pvalue of Z when X = 8. So

X = 12

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{12 - 12.6}{2.5}$

$Z = -0.24$

$Z = -0.24$ has a pvalue of 0.4052

X = 8

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{8 - 12.6}{2.5}$

$Z = -1.84$

$Z = -1.84$ has a pvalue of 0.0329

0.4052 - 0.0329 = 0.3723

37.23% probability that randomly selected homework will require between 8 and 12 minutes to grade

7 0

2 years ago

Use the expression (a+b)(c-d)+9e(a+b)(c−d)+9eleft parenthesis, a, plus, b, right parenthesis, left parenthesis, c, minus, d, rig

jek_recluse [69]

Answer:

Difference: ¨c-d¨ variable: ¨a¨ coefficient: ¨9¨

Step-by-step explanation:

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2 years ago

Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val

Cloud [144]

Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

$A(t) = Pe^{rt}$

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

$330 in an account in the year 1999

This means that $P = 330$

$590 in the year 2007

2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

$A(t) = Pe^{rt}$

$590 = 330e^{8r}$

$e^{8r} = \frac{590}{330}$

$e^{8r} = 1.79$

Applying ln to both sides:

$\ln{e^{8r}} = \ln{1.79}$

$8r = \ln{1.79}$

$r = \frac{\ln{1.79}}{8}$

$r = 0.0726$

Determine the value of the account, to the nearest dollar, in the year 2009.

2009 is 10 years after 1999, so this is A(10).

$A(t) = 330e^{0.0726t}$

$A(10) = 330e^{0.0726*10} = 682$

The value of the account in the year 2009 will be $682.

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2 years ago

Which of the x values are solutions to the inequality 4(2 – x) > –2x – 3(4x + 1)? Check all that apply.

steposvetlana [31]

4(2 - x) > -2x - 3(4x + 1)
8 - 4x > -2x - 12x - 3
-4x + 2x + 12x > -3 - 8
10x > -11
x > -11/10
x > -1.1

Therefore, x = 0 and x = 10 zre solutions to the inequality.

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2 years ago

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insens350 [35]

C .) The - 3 at the end of the function means the function is shifted down 3 units. This also shifts the asymptote down 3 units.

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2 years ago

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