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Andrei [34K]
3 years ago
7

In a shipment of 71 vials, only 13 do not have hairline cracks. If you randomly select 2 vials from the shipment, what is the pr

obability that none of the 2 vials have hairline cracks.
Mathematics
1 answer:
yanalaym [24]3 years ago
6 0

Answer:

0.0313

Step-by-step explanation:

Given that in the shipment of 71 vials, only 13 do not have hairline cracks

Probability of not having a hairline cracks = \frac{13}{71}

If you randomly select 2 vials from the shipment

(Probability of not having a hairline cracks)' = \frac{13-1}{71-1} = \frac{12}{70}

what is the probability that none of the 2 vials have hairline cracks.

i.e (\frac{13}{71}) * (\frac{12}{70})

= 0.183 × 0.171

= 0.0313

Hence, the probability that none of the 2 vials have hairline cracks = 0.0313

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How do you calculate a net worth statement?
marissa [1.9K]

Answer:

a. Assets - Liabilities

Step-by-step explanation:

Assets/Liabilities is a management method to minimize risk. So, it cannot be the answer.

There is no formula like "Assets + Liabilities" and "Assets x Liabilities" in accounting to calculate the net worth. Therefore, those can be eliminated.

We know that net worth is calculated by deducting all liabilities (long-term and short-term) from net assets. Therefore, option (a) is the correct answer.

8 0
3 years ago
Is it true that the planes x + 2y − 2z = 7 and x + 2y − 2z = −5 are two units away from the plane x + 2y − 2z = 1?
zhuklara [117]

Lets Find It Out..

First we'll find the equation of ALL planes parallel to the original one.

As a model consider this lesson:

Equation of a plane parallel to other

The normal vector is:
<span><span>→n</span>=<1,2−2></span>

The equation of the plane parallel to the original one passing through <span>P<span>(<span>x0</span>,<span>y0</span>,<span>z0</span>)</span></span>is:

<span><span>→n</span>⋅< x−<span>x0</span>,y−<span>y0</span>,z−<span>z0</span>>=0</span>
<span><1,2,−2>⋅<x−<span>x0</span>,y−<span>y0</span>,z−<span>z0</span>>=0</span>
<span>x−<span>x0</span>+2y−2<span>y0</span>−2z+2<span>z0</span>=0</span>
<span>x+2y−2z−<span>x0</span>−2<span>y0</span>+2<span>z0</span>=0</span>

Or

<span>x+2y−2z+d=0</span> [1]
where <span>a=1</span>, <span>b=2</span>, <span>c=−2</span> and <span>d=−<span>x0</span>−2<span>y0</span>+2<span>z0</span></span>

Now we'll find planes that obey the previous formula and at a distance of 2 units from a point in the original plane. (We should expect 2 results, one for each half-space delimited by the original plane.)
As a model consider this lesson:

Distance between 2 parallel planes

In the original plane let's choose a point.
For instance, when <span>x=0</span> and <span>y=0</span>:
<span>x+2y−2z=1</span> => <span>0+2⋅0−2z=1</span> => <span>z=−<span>12</span></span>
<span>→<span>P1</span><span>(0,0,−<span>12</span>)</span></span>

In the formula of the distance between a point and a plane (not any plane but a plane parallel to the original one, equation [1] ), keeping <span>D=2</span>, and d as d itself, we get:

<span><span>D=<span><span>|a<span>x1</span>+b<span>y1</span>+c<span>z1</span>+d|</span><span>√<span><span>a2</span>+<span>b2</span>+<span>c2</span></span></span></span></span>
<span>2=<span><span><span>∣∣</span>1⋅0+2⋅0+<span>(−2)</span>⋅<span>(−<span>12</span>)</span>+d<span>∣∣</span></span><span>√<span>1+4+4</span></span></span></span>
<span><span>|d+1|</span>=2⋅3</span> => <span><span>|d+1|</span>=6</span>First solution:
<span>d+1=6</span> => <span>d=5</span>
<span>→x+2y−2z+5=0</span>Second solution:
<span>d+1=−6</span> => <span>d=−7</span>
<span>→x+2y−2z−7=<span>0</span></span></span>
8 0
3 years ago
Paul spent $13 on writing utensils. Pens cost $3 each and pencils cost $1 each. If he bought a total of 7, how many of each did
dem82 [27]
The answer is 3 pens and 4 pencils. Hope this helps
3 0
2 years ago
Read 2 more answers
Simplify<br> 7^2x^-3y/49x^-3y^-2
olya-2409 [2.1K]
\frac{7^2x^{-3}}{49x^{-3} y^{-2}} \\ \\  \frac{49x^{-3}y}{49x^{-3}y^{-2}} \\ \\  \frac{49x^{-3}y}{49x^{-3} \times  \frac{1}{y^2} } \\ \\  \frac{49x^{-3}y}{ \frac{49x^{-3}}{y^2} } \\ \\ 49x^{-3} y \times  \frac{y^2}{49x^{-3}} \\ \\  \frac{x^{-3}}{x^{-3}} yy^2 \\ \\ 1 \times yy^2 \\ \\ yy^2 \\ \\ y^3 \\ \\

The answer is: y^3.
4 0
3 years ago
Read 2 more answers
A machine is designed to fill 16-ounce bottes of shampoo. When the machine is working properly, the amount poured into the bottl
vlada-n [284]

Answer:

(D) 15.90 to 16.20 ounces

Step-by-step explanation:

Confidence Interval = mean + or - (t×sd)/√n

Mean = 16.05 ounces, sd = 0.1 ounce, n = 4, degree of freedom = n - 1 = 4 - 1 = 3, t = 3.182

Lower limit = 16.05 - (3.182×0.1)/√4 = 16.05 - 0.15 = 15.90 ounces

Upper limit = 16.05 + (3.182×0.1)/√4 = 16.05 + 0.15 = 16.20 ounces

The sample mean will fall from 15.90 to 16.20 ounces

7 0
3 years ago
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