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2 years ago

6

Mai runs each lap in 8 minutes. She will run more than 88 minutes today. What are the possible numbers of laps she will run toda

y? Use n for the number of laps she will run today.

1 answer:

Readme [11.4K]2 years ago

4 0

Answer: 88<8n

Step-by-step explanation:

Because 88 is the least amount of time, it goes on one side of the inequity and the 8n goes on the other

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Given: ABCD trapezoid, BK ⊥ AD , AB=DC AB=8, AK=4 Find: m∠A, m∠B

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Answer:

$m\angle B=m\angle C=120^{\circ}$

$m\angle A=m\angle D=60^{\circ}$

Step-by-step explanation:

Trapezoid ABCD is isosceles trapezoid, because AB = CD (given). In isosceles trapezoid, angles adjacent to the bases are congruent, then

$\angle A\cong \angle D;$
$\angle B\cong \angle C.$

Since BK ⊥ AD, the triangle ABK is right triangle. In this triangle, AB = 8, AK = 4. Note that the hypotenuse AB is twice the leg AK:

$AB=2AK.$

If in the right triangle the hypotenuse is twice the leg, then the angle opposite to this leg is 30°, so,

$m\angle ABK=30^{\circ}$

Since BK ⊥ AD, then BK ⊥ BC and

$m\angle KBC=90^{\circ}$

Thus,

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2 years ago

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2 years ago

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Answer:

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Step-by-step explanation:

Getting all three marbles of green color only happens if every draw is a green marble. On the first marble draw, the urn has 10 marbles in it, out of which 5 are green. So the probability of drawing a green marble on this first draw is $\frac{5}{10}$

Then, once this has happened, the second draw also needs to be a green marble. At this point in the urn there are only 9 marbles left, and only 4 of them are green. So the probability of drawing a green marble at this point is $\frac{4}{9}$

Afterwards, on the last draw, a green marble also needs to be drawn. At this point there are only 8 marbles left on the urn, and only 3 of them are green. So the probability of drawing a green marble on this last draw is $\frac{3}{8}$

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3 years ago