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3 years ago

Line segment ON is perpendicular to line segment ML, and PN = 10.

Mathematics

2 answers:

nekit [7.7K]3 years ago

4 0

300 sq units. hope it will help you!

Ne4ueva [31]3 years ago

3 0

we know that

Triangle MOL is an isosceles triangle

because

$MO=OL=radius\ of\ circle=25\ units$

$ON=25\ units$

$ON=OP+PN\\ 25=OP+10\\ OP=25-10\\ OP=15\ units$

Find the base MP

Applying the Pythagorean Theorem in the right triangle MPO

$MO^{2} =MP^{2}+PO^{2}\\ MP^{2}=MO^{2}-PO^{2}\\ MP^{2}=25^{2}-15^{2}\\ MP^{2}=400\\ MP=20\ units$

$ML=2*MP\\ ML=2*20\\ ML=40\ units$

Find the area of triangle MOL

$A=\frac{1}{2}ML*PO\\ \\ A=\frac{1}{2}*40*15\\ \\ A=300\ unit^{2}$

therefore

the answer is

the area of triangle MOL is

$300$ square units

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Answer:

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4 years ago

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lutik1710 [3]

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4 years ago

The television show 50 Minutes has been successful for many years. That show recently had a share of 29, meaning that among the

BigorU [14]

Answer:

(a) P (none) = 0.005873

(b) P ( at least one household ) = 0.9941

(c) P ( at most one household ) = 0.9960

(d) not wrong

Step-by-step explanation:

Given data

probability P = 29% = 0.29

no of sample n = 15

tv show X = 50

to find out

the probability that none of the households

the probability that at least one household

the probability that at most one household

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solution

the probability that none of the households is tuned to 50 Minutes is

P (none) = 10C0 × $P^{0}$ × $(1 - 0.29)^{15}$

P (none) = $(0.71)^{15}$

P (none) = 0.005873

and the probability that at least one household is tuned to 50 Minutes is

P ( at least one household ) = 1 - P(none)

P ( at least one household ) = 1 - 0.005873

P ( at least one household ) = 0.9941

the probability that at most one household is tuned to 50 Minutes is

P ( at most one household ) = P (none) + P (x=1)

P ( at most one household ) =0.9941 + 10C1 × $(0.23)^{1}$ × $(0.71)^{14}$

P ( at most one household ) = 0.9960

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3 years ago

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3 years ago

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SVETLANKA909090 [29]

Answer:

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Step-by-step explanation:

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Standardization does this by transforming individual scores from different normal distributions to a common normal distribution with a known mean, standard deviation, and percentiles.

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3 0

3 years ago