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fredd [130]
3 years ago
9

A contractor purchases a shipment of 100 transistors. It is his policy to test 10 of these transistors and to keep the shipment

only if at least 9 of the 10 are in working condition. If the shipment contains 20 defective transistors, what is the probability it will be kept?
Mathematics
1 answer:
Serjik [45]3 years ago
4 0

Answer:

Hence the probability of the at least 9 of 10 in working condition is 0.3630492

Step-by-step explanation:

Given:

total transistors=100

defective=20

To Find:

P(X≥9)=P(X=9)+P(X=10)

Solution:

There  are 20 defective and 80 working transistors.

Probability of at least 9 of 10 should be working out 80 working transistors

is given by,

P(X≥9)=P(X=9)+P(X=10)

<em>{80C9 gives set of working transistor and 20C1 gives 20 defective transistor and 100C10 is combination of shipment of 10 transistors}</em>

P(X≥9)={80C9*20C(10-9)}/(100C10)+{80C10*20(10-10)}/(100C10)

<em>(Use the permutation and combination calculator)</em>

P(X≥9)=(231900297200*20/17310309456440)

+(1646492110120/17310309456440)

P(X≥9)=0.267933+0.0951162

P(X≥9)=0.3630492

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3 years ago
Find the area of the composite figure.
Roman55 [17]

Answer:

The Area of the composite figure would be 76.26 in^2

Step-by-step explanation:

<u>According to the Figure Given:</u>

Total Horizontal Distance = 14 in

Length = 6 in

<u>To Find :</u>

The Area of the composite figure

<u>Solution:</u>

Firstly we need to find the area of Rectangular part.

So We know that,

\boxed{ \rm \: Area  \:  of \:  Rectangle = Length×Breadth}

Here, Length is 6 in but the breadth is unknown.

To Find out the breadth, we’ll use this formula:

\boxed{\rm \: Breadth = total  \: distance - Radius}

According to the Figure, we can see one side of a rectangle and radius of the circle are common, hence,

\longrightarrow\rm \: Length \:  of \:  the  \: circle = Radius

  • Since Length = 6 in ;

\longrightarrow \rm \: 6 \: in   = radius

Hence Radius is 6 in.

So Substitute the value of Total distance and Radius:

  • Total Horizontal Distance= 14
  • Radius = 6

\longrightarrow\rm \: Breadth = 14-6

\longrightarrow\rm \: Breadth = 8 \: in

Hence, the Breadth is 8 in.

Then, Substitute the values of Length and Breadth in the formula of Rectangle :

  • Length = 6
  • Breadth = 8

\longrightarrow\rm \: Area \:  of  \: Rectangle = 6 \times 8

\longrightarrow \rm \: Area \:  of  \: Rectangle = 48 \: in {}^{2}

Then, We need to find the area of Quarter circle :

We know that,

\boxed{\rm Area_{(Quarter \; Circle) }  = \cfrac{\pi{r} {}^{2} }{4}}

Now Substitute their values:

  • r = radius = 6
  • π = 3.14

\longrightarrow\rm Area_{(Quarter \; Circle) } =  \cfrac{3.14 \times 6 {}^{2} }{4}

Solve it.

\longrightarrow\rm Area_{(Quarter \; Circle) } =  \cfrac{3.14 \times 36}{4}

\longrightarrow\rm Area_{(Quarter \; Circle) } =  \cfrac{3.14 \times \cancel{{36} } \: ^{9} }{ \cancel4}

\longrightarrow\rm Area_{(Quarter \; Circle)} =3.14 \times 9

\longrightarrow\rm Area_{(Quarter \; Circle) } = 28.26 \:  {in}^{2}

Now we can Find out the total Area of composite figure:

We know that,

\boxed{ \rm \: Area_{(Composite Figure)} =Area_{(rectangle)}+ Area_{ (Quarter Circle)}}

So Substitute their values:

  • \rm Area_{(rectangle)} = 48
  • \rm Area_{(Quarter Circle)} = 28.26

\longrightarrow \rm \: Area_{(Composite Figure)} =48 + 28 .26

Solve it.

\longrightarrow \rm \: Area_{(Composite Figure)} =\boxed{\tt 76.26 \:\rm in {}^{2}}

Hence, the area of the composite figure would be 76.26 in² or 76.26 sq. in.

\rule{225pt}{2pt}

I hope this helps!

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