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Vaselesa [24]
3 years ago
14

Given that a randomly chosen quadrilateral has four right angles, what is the probability that the quadrilateral also has four e

qual side lengths? Express your answer in percent form, round the the nearest whole percent

Mathematics
1 answer:
Alexxx [7]3 years ago
7 0

Answer:

0.25

Step-by-step explanation:

This is the question on conditional probability

Let A - the quadrilateral has four right angles

B - the quadrilateral has four equal side lengths

Required probability = P(B/A)

By definition of continuous probability

P(B/A) = \frac{P(AB)}{P(A)}

P(AB) = \frac{2}{20} =0.10

P(A) = \frac{8}{20} =0.4

Hence given probability = \frac{0.1}{0.4} =0.25

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Write this quadratic equation in standard form.
oksano4ka [1.4K]

Answer:

-x^{2} + 3x -8 = 0

Step-by-step explanation:

  1. x^{2} + x -8  - 2x = 0
  2. x^{2} + 3x -8 = 0
6 0
2 years ago
A survey of 100 high school students provided this
HACTEHA [7]

Answer:

35/100

Step-by-step explanation:

You have to find how many juniors there are, so if you add 13, 20, and 2 you get 35. For the denominator you have to find the total number of students, so just add all of the numbers together to get 100. So there is a 35/100 chance that a randomly selected student is a junior.

5 0
1 year ago
Plsss help!!!<br>y = ??<br>X =???​
Lyrx [107]

Answer:

x=80* y=20* and X=100*

Step-by-step explanation:

If we assume PQ || OR then by alternate interior angles x is equal to 80 degrees and y is equal to 20 degrees.

We know that a triangle adds up to 180 degrees.

Angle ROX and XRO add up to 100 degrees. Leaving RXO equal to 80 degrees. (180-100=80).

By opposite angles, QXP is equal to 80 degrees also.

Then we are left with solving for X.

Supplementary angles add up to 180 degrees.

Subtract angle OXR from 180 and we are left with angle OXP = X = 100 degrees.

8 0
3 years ago
Can anyone do these two<br> a)√3(√3-1)<br> b)√5(√10+√2)
gogolik [260]
A 1.267949192
b 10.23334547
7 0
3 years ago
A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is .5
Elodia [21]

Answer:

7.3% percentage of the bearings produced will not be acceptable.

Step-by-step explanation:

Consider the provided information.

Average diameter of the bearings it produces is .500 inches. A bearing is acceptable if its diameter is within .004 inches of this target value.

Let X is the normal random variable which represents the diameter of bearing.

Thus, 0.500-0.004<X<0.500+0.004

0.496<X<0.504

The bearings have normally distributed diameters with mean value .499 inches and standard deviation .002 inches.

Use the Z score formula: \frac{X-\mu}{\sigma}

Therefore

\frac{0.496-0.499}{0.002}\leq z\leq \frac{0.504-0.499}{0.002}

\frac{-0.003}{0.002}\leq z\leq \frac{0.005}{0.002}

-1.5\leq z\leq 2.5

Now use the standard normal table and determine the probability of that a ball bearing will be acceptable.

P(-1.5\leq z\leq 2.5)=0.9938-0.0668=0.9270

We need to find the percentage of the bearings produced will not be acceptable.

So subtract it from 1 as shown.

1-0.9270=0.073

Hence, 7.3% percentage of the bearings produced will not be acceptable.

3 0
3 years ago
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