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hjlf
1 year ago
13

Quadrilateral HGEF is a scaled copy of quadrilateral DCAB. What is themeasurement of lin EG?

Mathematics
1 answer:
lozanna [386]1 year ago
3 0

Answer:

14 units

Explanation:

If quadrilaterals HGEF and DCAB are similar, then the ratio of some corresponding sides is:

\frac{FH}{BD}=\frac{EG}{AC}

Substitute the given side lengths:

\begin{gathered} \frac{6}{3}=\frac{EG}{7} \\ 2=\frac{EG}{7} \\ \implies EG=2\times7 \\ EG=14 \end{gathered}

The measurement of line EG is 14 units.

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Setler79 [48]

The right answer is 18.8333333... Or 18 7.5/9

4 0
3 years ago
Alice stand at point A and looks at the top of a 17.8 m tree TB, such that her line of sight makes an angle 38° with the horizon
Nesterboy [21]

Answer:

Step-by-step explanation:

let the horizontal distance be x

(17.8-1.5)/x = tan 38

or x = 16.3/tan38

or x = 20.863

4 0
3 years ago
Suppose that you are in charge of evaluating teacher performance at a large elementary school. One tool you have for this evalua
Strike441 [17]

Answer:

a) Standard error = 2

b) Range = (76.08, 83.92)

c) P=0.69

d) Smaller

e) Greater

Step-by-step explanation:

a) When we have a sample taken out of the population, the standard error of the mean is calculated as:

\sigma_m=\dfrac{\sigma}{\sqrt{n}}=\dfrac{10}{\sqrt{25}}=\dfrac{10}{5}=2

where n is te sample size (n=25) and σ is the population standard deviation (σ=10).

Then, the standard error of the classroom average score is 2.

b) The calculations for this range are the same that for the confidence interval, with the difference that we know the population mean.

The population standard deviation is know and is σ=10.

The population mean is M=80.

The sample size is N=25.

The standard error of the mean is σM=2.

The z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

MOE=z\cdot \sigma_M=1.96 \cdot 2=3.92

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 80-3.92=76.08\\\\UL=M+t \cdot s_M = 80+3.92=83.92

The range that we expect the average classroom test score to fall 95% of the time is (76.08, 83.92).

c) We can calculate this by calculating the z-score of X=79.

z=\dfrac{X-\mu}{\sigma}=\dfrac{79-80}{2}=\dfrac{-1}{2}=-0.5

Then, the probability of getting a average score of 79 or higher is:

P(X>79)=P(z>-0.5)=0.69146

The approximate probability that a classroom will have an average test score of 79 or higher is 0.69.

d) If the sample is smaller, the standard error is bigger (as the square root of the sample size is in the denominator), so the spread of the probability distribution is more. This results then in a smaller probability for any range.

e) If the population standard deviation is smaller, the standard error for the sample (the classroom) become smaller too. This means that the values are more concentrated around the mean (less spread). This results in a higher probability for every range that include the mean.

6 0
3 years ago
Solve:
drek231 [11]

The solutions to the inequalities are x >1 and x < 6

<h3>How to solve the inequalities?</h3>

The inequality expression is given as:

-2x + 5 < 3x + 10

Collect the like terms in the above inequality

-2x - 3x < 10 - 5

Evaluate the like terms

-5x < 5

Divide by -5

x >1

Also, we have

5(x - 2) <3x + 2

Open the bracket

5x - 10 < 3x + 2

Evaluate the like terms

2x < 12

Divide by 2

x < 6

Hence, the solutions to the inequalities are x >1 and x < 6

Read more about inequalities at

brainly.com/question/24372553

#SPJ1

5 0
1 year ago
Tell which value of the variable is the solution of the equation.
vitfil [10]

Answer:

w = 5

Step-by-step explanation:

6w/6 = 30/6

= 5

6 0
2 years ago
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