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3 years ago

Identify the correct explanation for why the triangles are similar. Then find OP and MN.

Mathematics

2 answers:

STALIN [3.7K]3 years ago

6 0

Answer:

Therefore, the correct explanation is;

C. ∠L ≅ ∠L By Reflexive Prop. ≅.

Since $\overline {OP}$ ║ $\overline {MN}$ , ∠LOP ≅ ∠LMN by the Corr. ∠s Post. Therefore, ΔLOP ~ ΔLMN by AA ~. OP = 4 and MN = 8

Step-by-step explanation:

The given parameters are;

$\overline {LP}$ = 5

$\overline {NP}$ = 10

$\overline {OP}$ = x - 3

$\overline {MN}$ = x + 1

A two column proof is presented as follows;

Statement ${}$ Reason

∠L ≅ ∠L ${}$ By Reflexive property of congruency

$\overline {OP}$ ║ $\overline {MN}$ ${}$ Given

∠LOP ≅ ∠LMN ${}$ By the Corresponding angles Postulate

Therefore

ΔLOP ~ ΔLMN ${}$ By AA similarity Postulate

Where we have that ΔLOP and ΔLMN, we get;

$\overline {OP}$ / $\overline {MN}$ = $\overline {LP}$ / $\overline {NP}$ = 5/10 = 1/2

∴ (x - 3)/(x + 1) = 1/2

2·(x - 3) = 1·(x + 1)

2·x - 6 = x + 1

2·x - x = 1 + 6 = 7

x = 7

$\overline {OP}$ = (x - 3) = 7 - 3 = 4

$\overline {OP}$ = 4

$\overline {MN}$ = x + 1 = 7 + 1 = 8

$\overline {MN}$ = 8

Therefore, the correct option is ∠L ≅ ∠L By Reflexive Prop. ≅.

Since $\overline {OP}$ ║ $\overline {MN}$ , ∠LOP ≅ ∠LMN by the Corr. ∠s Post. Therefore, ΔLOP ~ ΔLMN by AA ~. OP = 4 and MN = 8.

KatRina [158]3 years ago

5 0

Answer:

∠L≅∠L by the Reflex Prop. of ≅.

Since OP⎯⎯⎯⎯⎯∥MN⎯⎯⎯⎯⎯⎯,∠LOP≅∠LMN by the

Corr. ∠s Post. Therefore, △LOP∼△LMN

by AA∼. OP=2 and MN=6.

Step-by-step explanation:

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Step-by-step explanation:

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A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$