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Doss [256]
3 years ago
15

A shipment of 12 microwave ovens contains 4 defective units. A restaurant buys three of these units. What is the probability of

the restaurant buying at least two nondefective​ units?
Mathematics
1 answer:
hammer [34]3 years ago
5 0
The <u>correct answer</u> is:

20/27 = 0.7407.

Explanation:

To find the probability that the restaurant buys at least 2 non-defective units, we find the probability that they purchase either 2 or 3 non-defective units:

P(r) = _nC_r(p)^r(1-p)^{n-r}&#10;

They purchase <u>3 units total</u>, so this is <u>n</u>.
We want the <u>probability of either 2 or 3</u> non-defective units; this means <u>r is 2 and r is 3</u>.
In this case we want the probability that the units are non-defective.  Since there are 4 defective units, this means there are 12-4=8 non-defective units; this makes the <u>probability of a non-defective unit 8/12 = 2/3.  This is p</u>.

Using these, we have:
P(X=2 or X=3)=_3C_2(\frac{2}{3})^2(1-\frac{2}{3})^{3-2}+_3C_3(\frac{2}{3})^3(1-\frac{2}{3})^{3-3}&#10;\\&#10;\\=\frac{3!}{2!1!}(\frac{2}{3})^2(\frac{1}{3})^1+\frac{3!}{3!0!}(\frac{2}{3})^3(\frac{2}{3})^0&#10;\\&#10;\\=3(\frac{4}{9})(\frac{1}{3})+1(\frac{8}{27})(1)&#10;\\&#10;\\=\frac{12}{27}+\frac{8}{27}=\frac{20}{27}=0.7407
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2 years ago
Find the factorial form<br> 3x2x1
Lubov Fominskaja [6]

Answer:

3x2x1=3!

Step-by-step explanation:

using nx(n-1)x_x1=n

3x2x1=3!

hence,3x2x1=3!

8 0
3 years ago
2<br>If A=<br>4 3<br>find A1 using<br>elementary row operations.​
MA_775_DIABLO [31]

Answer:   A^{-1}=\left[\begin{array}{cc}\frac{3}{2}&-\frac{1}{2}\\-2&1\end{array}\right]

<u>Step-by-step explanation:</u>

                  \left[\begin{array}{cc}2&1\\4&3\end{array}\right]=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]

\dfrac{1}{2}Row\ 1\rightarrow\left[\begin{array}{cc}1&\frac{1}{2}\\4&3\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}&0\\0&1\end{array}\right]

Row\ 2 -4 \ Row\ 1\rightarrow \left[\begin{array}{cc}1&\frac{1}{2}\\0&1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}&0\\-2&1\end{array}\right]

Row\ 1-\dfrac{1}{2}\ Row\ 2 \rightarrow \left[\begin{array}{cc}1&0\\0&1\end{array}\right]=\left[\begin{array}{cc}\frac{3}{2}&-\frac{1}{2}\\-2&1\end{array}\right]

5 0
3 years ago
4 - x + <img src="https://tex.z-dn.net/?f=6%5E%7B2%7D" id="TexFormula1" title="6^{2}" alt="6^{2}" align="absmiddle" class="latex
Lemur [1.5K]

Answer:

x ≤ 19

Step-by-step explanation:

The instructions here are probably "solve for x."  Please include them.

4 - x + 6^2 ≥  21

becomes 4 - x + 36 ≥  21

Now combine like terms.  4 and 36 combine to 40:  40 - x ≥ 21, and so:

19 - x ≥ 0

Adding x to both sies results in

x ≤ 19

Please, include the instructions when you post a question.  Thanks.

3 0
2 years ago
NO LINKS!!! Part 2: Find the Lateral Area, Total Surface Area, and Volume. Round your answer to two decimal places.​
slava [35]

Answer:

<h3><u>Question 7</u></h3>

<u>Lateral Surface Area</u>

The bases of a triangular prism are the triangles.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the triangles (bases).

\implies \sf L.A.=2(10 \times 6)+(3 \times 6)=138\:\:m^2

<u>Total Surface Area</u>

Area of the isosceles triangle:

\implies \sf A=\dfrac{1}{2}\times base \times height=\dfrac{1}{2}\cdot3 \cdot \sqrt{10^2-1.5^2}=\dfrac{3\sqrt{391}}{4}\:m^2

Total surface area:

\implies \sf T.A.=2\:bases+L.A.=2\left(\dfrac{3\sqrt{391}}{4}\right)+138=167.66\:\:m^2\:(2\:d.p.)

<u>Volume</u>

\sf \implies Vol.=area\:of\:base \times height=\left(\dfrac{3\sqrt{391}}{4}\right) \times 6=88.98\:\:m^3\:(2\:d.p.)

<h3><u>Question 8</u></h3>

<u>Lateral Surface Area</u>

The bases of a hexagonal prism are the pentagons.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the pentagons (bases).

\implies \sf L.A.=5(5 \times 6)=150\:\:cm^2

<u>Total Surface Area</u>

Area of a pentagon:

\sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}a^2

where a is the side length.

Therefore:

\implies \sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}\cdot 5^2=43.01\:\:cm^2\:(2\:d.p.)

Total surface area:

\sf \implies T.A.=2\:bases+L.A.=2(43.01)+150=236.02\:\:cm^2\:(2\:d.p.)

<u>Volume</u>

\sf \implies Vol.=area\:of\:base \times height=43.011193... \times 6=258.07\:\:cm^3\:(2\:d.p.)

8 0
1 year ago
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