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beks73 [17]
3 years ago
7

Use quadrilateral ABCD to find the value of X. The figure is not drawn to scale. Use the following dimensions: mABC=4x, mBCD=3x,

mCDA=2x, mDAB=3x
Find the measure of each angle:

mABC=___ mBCD=___ mCDA=___ mDAB=___

Mathematics
2 answers:
babymother [125]3 years ago
8 0

Answer:

The answer to your question is:

mABC = 120°               mBCD = 90°        mCDA = 60°          mDAB = 90°

Step-by-step explanation:

To solve this problem, we remember that the sum of the angles in a quadrilateral = 360°.

Then

360° = mABC +  mBCD + mCDA  + mDAB

360 = 3x + 4x + 3x + 2x                   substitution

360 = 12x                                         simplifying

x = 360/12

x = 30                                          

Now, we find the values of each angle

mABC = 4(30) = 120°         mBCD = 3(30) = 90°       mCDA = 2(30) = 60°

mDAB = 3(30) = 90°

Dmitrij [34]3 years ago
5 0

Answer: x=30

Step-by-step explanation:

The sum of the angles must be 360.

4x+3x+3x+2x = 12x

12x = 360

x =360/12 = 30

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The concentration of particles in a suspension is 50 per mL. A 5 mL volume of the suspension is withdrawn. a. What is the probab
kolezko [41]

Answer:

(a) 0.6579

(b) 0.2961

(c) 0.3108

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Step-by-step explanation:

The random variable <em>X</em> can be defined as the number of particles in a suspension.

The concentration of particles in a suspension is 50 per ml.

Then in 5 mL volume of the suspension the number of particles will be,

5 × 50 = 250.

The random variable <em>X</em> thus follows a Poisson distribution with parameter, <em>λ</em> = 250.

The Poisson distribution with parameter λ, can be approximated by the Normal distribution, when λ is large say λ > 10.  

The mean of the approximated distribution of X is:

μ = λ  = 250

The standard deviation of the approximated distribution of X is:

σ = √λ  = √250 = 15.8114

Thus, X\sim N(250, 250)

(a)

Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows:

P(235

                             =P(-0.95

Thus, the value of P (235 < <em>X</em> < 265) = 0.6579.

(b)

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

P(48

                             =P(-0.28

Thus, the value of P(48.

(c)

A 10 mL sample is withdrawn.

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

P(48

                             =P(-0.40

Thus, the value of P(48.

(d)

Let the sample size be <em>n</em>.

P(48

                             0.95=P(-z

The value of <em>z</em> for this probability is,

<em>z</em> = 1.96

Compute the value of <em>n</em> as follows:

z=\frac{\bar X-\mu}{\sigma/\sqrt{n}}\\\\1.96=\frac{48-50}{15.8114/\sqrt{n}}\\\\n=[\frac{1.96\times 15.8114}{48-50}]^{2}\\\\n=240.1004\\\\n\approx 241

Thus, the sample selected must be of size 240.

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