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2 years ago

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In the diagram, mFLI is 106°, mFLG = (2x – 1)°, mGLH = (x + 17)°, and mHLI = (4x – 15)°. What is the measure of the smallest ang

le in the diagram? 15° 29° 32° 45°

2 answers:

klio [65]2 years ago

4 0

Ok I think I have a visual, but if this isn't want the question wants, don't blame me because you couldn't put a diagram up.

106 = (2x-1) + (x+17)+ (4x-15)
106 = 7x + 1
105 = 7x
/7 /7
15 = x

mFLG = 2x - 1
2(15) - 1
30 - 1
29

mGLH = x+17
15 + 17
32

mHLI = 4x - 15
4(15) - 15
60 - 15
45

29 < 32
29 < 45

Therefore, the smallest angle is 29.

Proof = 29 + 32 + 45 = 106.

Kipish [7]2 years ago

4 0

Answer:

29 degrees

Step-by-step explanation:

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Answer:

0% probability that at least 1,500 will agree to respond

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

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Can be approximated to a normal distribution, using the expected value and the standard deviation.

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The standard deviation of the binomial distribution is:

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Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

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$n = 15000, p = 0.09$

So

$\mu = E(X) = np = 15000*0.09 = 1350$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{15000*0.09*0.91} = 35.05$

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6 0

2 years ago

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

Ne4ueva [31]

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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So, $P\left(\frac{E}{A}\right)=0.02$

Now, from the conditional probability,

$P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}$

$\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)$

$\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)$

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$P\left(\frac{E}{A}\right)=0.40$

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$P\left(\frac{E}{A}\right)=0.98$

and $\Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)$

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(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

$=P(E\cap C)$

$= 0.0686$ [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

$P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)$

$\Rightarrow P(E)=0.018+0.012+0.0686$

$\Rightarrow P(E)=0.0986$

So, the total probability that the part produced get shipped

$=1-P(E)=1-0.0986=0.9014$

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$=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)$

$=(0.9-0.018)+(0.03-0.012)$

$=0.9$

So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

$=\frac{\text{Probabilily that shipped part is 'good'}}{\text{Probability of total shipped parts}}$

$=\frac{0.9}{0.9014}$

$=0.9984$

(c) The probability that the 'bad' (defective} parts get shipped

=1- the probability that the 'good' parts get shipped

=1-0.9984

=0.0016

5 0

2 years ago