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2 years ago

Pls help! In the figure below, m∠ABC = 57 degrees and m∠DAB =116 degrees. What is m∠ABC?

Mathematics

2 answers:

ICE Princess25 [194]2 years ago

8 0

Answer:

59 degrees

Step-by-step explanation:

you can use angle DAB to find out angle BAC by subtracting 180 and 116 to get 64

add 57 and 64 to get 121

subtract 180 and 121 to get 59 degrees.

tangare [24]2 years ago

4 0

Answer:

59°

Step-by-step explanation:

116-57=59

the outside angle is the 2 inner angles added together

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As we can see that the sides of the rectangle have been doubled

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So perimeter of new rectangle = 2 (16 ) = 32 feet

Option C is correct

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2 years ago

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Umnica [9.8K]

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Step-by-step explanation:

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2 years ago

What is 14/15 over 4/7 equal to

Aneli [31]

Answer:

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Step-by-step explanation:

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2 years ago

Find the median 13,22,21,25,15

Ber [7]

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3 years ago

Read 2 more answers

Suppose you do not know the population mean fee charged to H&R Block customers last year. Instead, suppose you take a sample

puteri [66]

Answer:

i $\to$ a

$n = 96040000$

i $\to$ b

$n_1 =24010000$

i $\to$ c

$n_2 =41602500$

ii $\to$ a

$E = 58.16$

ii $\to$ b

$291.84 < \mu < 408.16$ \

ii $\to$ c

There is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval

Step-by-step explanation:

From the question we are told that

The sample size is $n = 8$

The sample mean is $\= x = \$ 350$

The sample standard deviation is $\$ 100$

Considering question i

i $\to$ a

At $E = 0.02$

given that the confidence level is 95% = 0.95

the level of significance would be $\alpha =1-0.95 = 0.05$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{ \alpha }{2} } = 1.96$

So the sample size is mathematically evaluated as

$n = [ \frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ]^2$

=> $n =[ \frac{ 1.96 * 100}{ 0.02} ]^2$

=> $n = 96040000$

i $\to$ b

At $E_1 = 0.04$ and confidence level = 95% => $\alpha_1 = 0.05$ => $Z_{\frac{\alpha_1 }{2} } = 1.96$

$n_1 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_1} ]^2$

=> $n_1 =[ \frac{ 1.96 * 100}{ 0.04} ]^2$

=> $n_1 =24010000$

i $\to$ c

At $E_2 = 0.04$ confidence level = 99% => $\alpha_2 = 0.01$

The critical value of $\frac{\alpha_2 }{2}$ from the normal distribution table is

$Z_{\frac{ \alpha_2 }{2} } = 2.58$

=> $n_2 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_2} ]^2$

=> $n_2 =[ \frac{ 2.58 * 100}{ 0.04} ]^2$

=> $n_2 =41602500$

Considering ii

Given that the level of significance is $\alpha = 0.10$

Then the critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

substituting values

$E = 1.645 * \frac{100 }{\sqrt{8} }$

$E = 58.16$

Generally the 90% confidence interval is mathematically evaluated as

$\= x - E < \mu < \= x + E$

=> $350 - 58.16 < \mu < 350 + 58.16$

=> $291.84 < \mu < 408.16$

So the interpretation is that there is 90% confidence that the mean fee charged to H&R Block customers last year is in the interval .So there is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval.

8 0

2 years ago