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3 years ago

Is ∆LMP≅∆NMP by HL? If so, which legs allow the use of HL?

Mathematics

1 answer:

Westkost [7]3 years ago

7 0

Answer:

Step-by-step explanation:

For the 2 triangles to be ≅by HL we need the hypothenuse and pair of legs to be congruent.

We know for sure that we have a pair of legs ≅

the legs MP≅MP by reflexive propriety ( is congruent with itself)

But we do not know from the picture that the hypothenuses are ≅

in ΔLMP the hypothenuse is LM

inΔNMP the hypothenuse is NM

So unless is given that LM si congruent to MN the Δs are not ≅ by HL theorem

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A manufacturing facility is responsible for producing two components that are produced at states 1 (poor), 2 (below average), 3

garik1379 [7]

Answer:

a) i) P( final product is acceptable ) = 3/5

ii) P ( final product not acceptable ) = 2/5

B) P ( final product is acceptable | X =2 ) = 2/5

C) P ( Y = 3 | Final product is acceptable ) = 2/5

Step-by-step explanation:

Given data :

5 states : 1 (poor), 2 (below average), 3 (average), 4 (good), and 5 (excellent)

Number of components produced = 2

X ( component 1 ) ∴ P(X) = 1/5

Y ( component 2 ) ∴ P(Y) = 1/5

Aim : Sum of Final product needs to be ≥ 6

<u>A) Probability of a final product been acceptable and also unacceptable</u>

P( final product is acceptable ) ; P(X + Y ≥ 6 )

= 1/5 ∑ P( Y ≥ 6 - x )

= 1/5 ( P( Y ≥ 5 ) + P( Y ≥ 4 ) + P( Y ≥ 3 ) + P( Y ≥ 2 ) + P( Y ≥ 1 )

= 1/5 ( 1/5 + 2/5 + 3/5 +4/5 + 1 )

= 3/5

P ( final product not acceptable ) = 1 - 3/5 = 2/5

B) P ( final product is acceptable | X =2 )

∴ P ( Y ≥ 4 )

= 2/5

C) P ( Y = 3 | Final product is acceptable )

∴ P ( X ≥ 3 ) ≠ P( Y ≥ 3 ) ( because they are not independent )

= 1 - 3/5 = 2/5.

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3 years ago

Felicity set the thermostat in her living room to 68°F. The room temperature t, in degrees Fahrenheit, m minutes after the therm

Ahat [919]

Answer with explanation:

Temperature in the living room =68°F

The equation which satisfies ,the room temperature t, in degrees Fahrenheit, m minutes after the thermostat is activated, satisfies

t=2 Cos (0.21 m) +68----------(1)

⇒To determine the period of the function

Z= A cos (sx+q)+r

As period of cos x is 2 π.

So,the given function has period

$=\frac{2\pi}{s}$

So, the period of function 1, is given by

$=\frac{2\pi}{0.21}$

⇒The meaning of period of the function is that after every period of $=\frac{2\pi}{0.21}$ the temperature of the room increases or decreases by an integer equal to 2.

⇒Cosine function has maximum value of 1 , and Minimum value of -1.

-1 ≤ Cos x ≤ 1

So, the Maximum value of function = 2 ×1+68°=70°----Maximum Temperature

And, the minimum value of function = 2 ×(- 1)+68°=66°----Minimum Temperature

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3 years ago