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4 years ago

If line AD is a tangent to circle B at point C, and m ABC = 55°, what is the measure of BAD?

Mathematics

1 answer:

nikdorinn [45]4 years ago

8 0

Answer:

145°

Step-by-step explanation:

There are a couple of ways you can get there:

1. ∠ACB is a right angle, 90°. Hence, ∠BAC is the complement of ∠ABC, so is ...

... ∠BAC = 90° -∠ABC = 90° -55° = 35°

Then, ∠BAC and ∠BAD are a linear pair, so total 180°. That makes ∠BAD the supplement of ∠BAC, so ...

... ∠BAD = 180° -35° = 145°

2. ∠BAD is the exterior angle at A for the triangle ABC. It will have a measure that is the sum of the opposite interior angles: given ∠ABC = 55° and right angle ACB = 90°.

... ∠BAD = 55° +90° = 145°

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3 years ago

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Barry Allen's older brother is only 3 years more than triple Barry's age. If the sum of their ages is 28, find their ages

Sloan [31]

B and a
barry=b
a=older bro

a is 3 more than 3b
a=3+3b
sum is 28
a+b=28
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3 years ago

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Refer to the random sample of customer order totals with an average of $78.25 and a population standard deviation of $22.50. a.

zysi [14]

Answer:

a) $78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

b) $78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

c) For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

d) $ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

Step-by-step explanation:

Part a

For this case we have the following data given

$\bar X = 78.25$ represent the sample mean for the customer order totals

$\sigma =22.50$ represent the population deviation

$n= 40$ represent the sample size selected

The confidence level is 90% or 0.90 and the significance level would be $\alpha=0.1$ and $\alpha/2 = 0.05$ and the critical value from the normal standard distirbution would be given by:

$z_{\alpha/2}=1.64$

And the confidence interval is given by:

$\bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

Part b

The sample size is now n = 75, but the same confidence so the new interval would be:

$78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

Part c

For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

Part d

The margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

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MatroZZZ [7]

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4 years ago

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Liula [17]

Answer:

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Step-by-step explanation:

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