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2 years ago

1. Determining the probability of a coin landing heads up by actually tossing a coin several times is an example of what?

Mathematics

2 answers:

Eddi Din [679]2 years ago

5 0

The correct answer is C

Oksana_A [137]2 years ago

4 0

Answer:

Step-by-step explanation:

Theoretical probability and Experimental probability are related in that theoretical probability is based on your reasoning (e.g. two sides; therefore, there is a 50% chance of heads), but experimental probability is where your numbers are based off of actual results (e.g. you landed head 1/4th of the time; therefore, there is a 25% chance of getting heads.)

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2115-0830 in millitary time please and thank you so much if i get it right

neonofarm [45]

Answer:

2115 = 9:15 PM

0830 = 8:30 AM

Step-by-step explanation:

12 - hour conversion.

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3 years ago

Which expressions listed on the left are equivalent to √36a^8/225a^2? Check all that apply. (Assume that a ≠ 0.)

MAVERICK [17]

Answer: ABE

Step-by-step explanation:

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3 years ago

Given: KL ║ NM , LM = 45, m∠M = 50° KN ⊥ NM , NL ⊥ LM Find: KN and KL

Mice21 [21]

Answer:

$KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08\\ \\KN=45\sin 50^{\circ}\approx 34.47$

Step-by-step explanation:

Given:

KL ║ NM ,

LM = 45

m∠M = 50°

KN ⊥ NM

NL ⊥ LM

Find: KN and KL

1. Consider triangle NLM. This is a right triangle, because NL ⊥ LM. In this triangle,

LM = 45

m∠M = 50°

So,

$\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}$

Also

$m\angle LNM=90^{\circ}-50^{\circ}=40^{\circ}$ (angles LNM and M are complementary).

2. Consider triangle NKL. This is a right triangle, because KN ⊥ NM . In this triangle,

$NL=45\tan 50^{\circ}$

$m\angle KLN=m\angle LNM=40^{\circ}$ (alternate interior angles)

$m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ}$ (angles KNL and KLN are complementary).

So,

$\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08$

and

$\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47$