Which one is the greatest?

In the figure shown, suppose m angle ABC=n and m angle ABD=2(m angle DBC). The angle bisector of angle DBC is BE. What is m angl

Yanka [14]

Answer:

n/6

Step-by-step explanation:

If angle bisector of angle DBC is BE. What is m angle EBC, then <DBE = <EBC

Since <ABC = <ABD+<DBC

n = 2<DBC+<DBC

n = 3<DBC

<DBC = n/3

Since <DBC = 2EBC

n/3 = 2<EBC

n/6 = <EBC

Hence <EBC = n/6

4 0

3 years ago

1 point

swat32

Answer:

$\log_{10}(147) = 2.1673$

Step-by-step explanation:

Given

$\log_{10} 3 = 0.4771$

$\log_{10} 5 = 0.6990$

$\log_{10} 7= 0.8451$

$\log_{10} 11 = 1.0414$

Required

Evaluate $\log_{10}(147)$

Expand

$\log_{10}(147) = \log_{10}(49 * 3)$

Further expand

$\log_{10}(147) = \log_{10}(7 * 7 * 3)$

Apply product rule of logarithm

$\log_{10}(147) = \log_{10}(7) + \log_{10}(7) + \log_{10}(3)$

Substitute values for log(7) and log(3)

$\log_{10}(147) = 0.8451 + 0.8451 + 0.4771$

$\log_{10}(147) = 2.1673$

3 0

2 years ago

The table and the graph below each show a different relationship between the same two variables, x and y:

Angelina_Jolie [31]

Answer:

The answer is A.

Step-by-step explanation:

thats what i think

7 0

3 years ago

The area of rectangle ABCD is 28 cm^2

Sonbull [250]

The length of AB is (x + 4) cm and the values of a and be are a = 10, b = 4

A rectangle is a quadrilateral in which opposite sides are equal and parallel to each other. The area of a rectangle is:

Area = length * width

From the image:

Length of AB = x + 4

Length of BC = x + 6

The area of rectangle ABCD = Length of AB * Length of BC

28 = (x + 4)(x + 6)

x² + 10x + 24 = 28

x² + 10x = 4

Comparing with x² + ax = b gives:

a = 10, b = 4

Therefore the length of AB is (x + 4) cm and the values of a and be are a = 10, b = 4

Find out more at: brainly.com/question/15019502

8 0

2 years ago

Do teachers find their work rewarding and satisfying? An article reports the results of a survey of 397 elementary school teache

Tju [1.3M]

Answer:

The 95% confidence interval would be given (0.0109;0.1651).

We are confident at 95% that the difference between the two proportions is between $0.0109 \leq p_{elementary} -p_{High school} \leq 0.1651$

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_A$ represent the real population proportion for elementary school

$\hat p_A =\frac{226}{397}=0.569$ represent the estimated proportion for elementary school

$n_A=397$ is the sample size required for Brand A

$p_B$ represent the real population proportion for high school teachers

$\hat p_B =\frac{129}{268}=0.481$ represent the estimated proportion for high school teachers

$n_B=268$ is the sample size required for Brand B

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$(0.569-0.481) - 1.96 \sqrt{\frac{0.569(1-0.569)}{397} +\frac{0.481(1-0.481)}{268}}=0.0109$

$(0.569-0.481) + 1.96 \sqrt{\frac{0.569(1-0.569)}{397} +\frac{0.481(1-0.481)}{268}}=0.1651$

And the 95% confidence interval would be given (0.0109;0.1651).

We are confident at 95% that the difference between the two proportions is between $0.0109 \leq p_{elementary} -p_{High school} \leq 0.1651$

5 0

2 years ago