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3 years ago

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Help Me PLZUse the diagram below to answer the questions that follow. Lines a and b are parallel, LaǁLb. Point L is midpoint of

segment EF. If the measure of ∠DFH is 98° , what is the measure of ∠DFO?∠OFL? ∠LFH?

1 answer:

horrorfan [7]3 years ago

7 0

Answer:

Use the transversal

Step-by-step explanation:

Line/transversal DG splits everything into 180° sections. Since angle DFH is 98°, DFO is 180°-98°. OFL is 98° since it is a transversal split. LFH is 180°-98°.

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Hi could some PLEASE help me !!!!

Nataliya [291]

Answer:

3x=78+x-2

Step-by-step explanation:

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3 years ago

√2.35^2 + √-4.27^2

Ostrovityanka [42]

Square roots and squares cancel eachother. therefore the problem is simply:
2.35 + -4.27= -1.92

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4 years ago

How to simplify 45/100

fredd [130]

-- Look for a common factor of both the top and bottom numbers.

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-- Divide the top and bottom numbers by their common factor.

   ==> 45/100 = 9/20 .

-- Look for another common factor of both the top and bottom numbers.

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3 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

irakobra [83]

Answer:

We need a sample size of at least 383.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

85% confidence level

So $\alpha = 0.15$ , z is the value of Z that has a pvalue of $1 - \frac{0.15}{2} = 0.925$ , so $Z = 1.44$ .

How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 85% confidence level with an error of at most 0.03

We need a sample size of at least n.

n is found with $M = 0.03, \pi = 0.21$

Then

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.44\sqrt{\frac{0.21*0.79}{n}}$

$0.03\sqrt{n} = 1.44\sqrt{0.21*0.79}$

$\sqrt{n} = \frac{1.44\sqrt{0.21*0.79}}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.44\sqrt{0.21*0.79}}{0.03})^{2}$

$n = 382.23$

Rounding up

We need a sample size of at least 383.

6 0

3 years ago

Mary drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 7 hours. When Mary drove h

ELEN [110]

There are two separate equations for this question that use the same variable.

Mary took 7 hours to complete a trip with traffic. This would use the following expression:

$7x$

Mary took 4 hours to complete the same trip without traffic. The average rate of speed was 27 mph faster than the trip with traffic. This uses the following expression:

$4(x+27)$

Both trips had the same distance. From the information provided, we can set up the following equation:

$7x = 4(x+27)$

x represents the average rate of speed on both trips.

Distribute 4 to each term in parentheses:

$4 \times x = 4x$
$4 \times 27 = 108$

$7x = 4x + 108$

Subtract 4x from both sides:

$3x = 108$

Divide both sides by 3 to get x by itself:

$x = 36$

The average speed for the slower trip is 36 mph.

We can plug this value into the first equation:

$7(36) = 7 \times 36 = 252$

Mary lives 252 miles away from the mountains.

7 0

3 years ago