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3 years ago

In the figure below, AC and BE are diameters of circle P.

Mathematics

2 answers:

scoundrel [369]3 years ago

6 0

Answer:

$arc\ DE=39^o$

Step-by-step explanation:

step 1

Find the value of w

we know that

$m\angle APB+m\angle APE=180^o$ ---> by supplementary angles (form a linear pair)

substitute the given values

$(4w+8)^o+(4w+4)^o=180^o$

$8w=180-12\\8w=168\\w=21$

step 2

Find the measure of angle EPD

we know that

$m\angle APB=m\angle EPD+m\angle DPC$ ----> by vertical angles

substitute

$(4w+8)^o=m\angle EPD+(2w+11)^o$

substitute the value of w

$(4(21)+8)^o=m\angle EPD+(2(21)+11)^o$

$92^o=m\angle EPD+53^o$

$m\angle EPD=92^o-53^o=39^o$

step 3

Find the measure of arc DE

we know that

$arc\ DE=m\angle EPD$ ----> by central angle

therefore

$arc\ DE=39^o$

Amanda [17]3 years ago

3 0

Answer:

Step-by-step explanation:

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3 years ago

You measure 50 textbooks' weights, and find they have a mean weight of 37 ounces. Assume the population standard deviation is 5.

Hoochie [10]

Answer:

90% confidence interval for the true population mean textbook weight is [35.79 ounces , 38.21 ounces].

Step-by-step explanation:

We are given that you measure 50 textbooks' weights, and find they have a mean weight of 37 ounces.

Assume the population standard deviation is 5.2 ounces.

Firstly, the Pivotal quantity for 90% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean weight = 37 ounces

$\sigma$ = population standard deviation = 5.2 ounces

n = sample of textbooks = 50

$\mu$ = true population mean textbook weight

<em>Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about population standard deviation.</em>

<u></u>

<u>So, 90% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5%

level of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X -1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X +1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

<u>90% confidence interval for</u> $\mu$ = [ $\bar X -1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X +1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $37-1.645 \times {\frac{5.2}{\sqrt{50} } }$ , $37+1.645 \times {\frac{5.2}{\sqrt{50} } }$ ]