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lakkis [162]
3 years ago
9

Andy claims that his bird can do single-digit multiplication. He shows the bird 12 cards with simple multiplication problems tha

t are either correct or incorrect. The bird correctly identifies the correctness of the card 9 times out of 12. If the bird were simply guessing, how many would the bird be expected to get correct? 3 9 6
Mathematics
2 answers:
kondaur [170]3 years ago
8 0

Answer:

The expected number of cards is 6.

Step-by-step explanation:

We know that,

'Expected value = Total number of trials × Probability of each trial'

Let C is the event of correct cards and C' is the event of incorrect cards.

Given,

Total number of cards = 12

Also, the correctness of the card is 9 times out of 12,

Thus, the number of correct cards,

n(C) = 9,

⇒ The number of incorrect cards,

n(C') = 12 - 9 = 3

Also, A card can be can be correct or incorrect,

Thus, The probability of correct card,

P(C)=\frac{1}{2}

And, the probability of incorrect card,

P(C')=\frac{1}{2}

Thus, the expected number of cards = expected number of correct card + expected number of incorrect card

= n(C) × P(C) + n(C') × P(C')

=9\times \frac{1}{2}+3\times \frac{1}{2}

=\frac{9}{2}+\frac{3}{2}

=\frac{12}{2}=6

Ray Of Light [21]3 years ago
6 0
I believe the answer is 6
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murzikaleks [220]

Answer:

The values for Fx(1,2) and Fy(1,2) are 5 and 2 respectively.

Approximation at points (1.1,1.9) is 0.7

Step-by-step explanation:

Given:

Tangent plane to  a surface z=5x+2y-10 as the function at point (1,2)

To find :

f(x,y) at (1,2)

partial derivatives of function w.r.t. (x and y) and value of that function at given points.

Solution:(refer the attachment also)

Now we know that

the equation of tangent plane at given points to the surface is given by,

f(x1,y1,z1) and z=f(x,y)

z-z1=Fx(x1,y1)*(x-x1)+Fy(x1,y1)*(y-y1)

here Fx(x1,y1) and Fy(x1,y1) are the partial derivatives of x and y.

now

taking partial derivative w.r.t. x we get

Fx(x1`,y1)=\frac{d}{dx} (5x+2y-10)

=5.

Then w.r.t y we get

Fy(x1,y1)=

\frac{d}{dy}(5x+2y-10)

=2.

The values for Fx(1,2) and Fy(1,2) are 5 and 2 respectively.

Using the Linearization or linear approximation we get

L(x,y)=f(x1,y1)+Fx(x,y)*(x-x1)+Fy(x,y)(y-y1)

=-1+5(x-1)+2(y-2)

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Approximation at F(1.1,1.9)

=5(1.1)+2(1.9)-10

=5.5+3.8-10

=0.7

Approximation at points (1.1,1.9) is 0.7

6 0
3 years ago
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