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Wewaii [24]
3 years ago
6

I don't know why I can't think (maybe because its late) but can someone answer this along with a clear explanation? Thanks

Mathematics
1 answer:
MrMuchimi3 years ago
5 0
A=(0,1); \ \ \ B=(1,0)


y = ax+b


\pmb{1^{o}} \\  \\  \begin{cases} 1 = 0*a + b  \\  0 = 1*a + b  \end{cases}   \\  \\ \begin{cases}    b = 1 \\ a+b = 0\end{cases}   \\  \\ \begin{cases}  b = 1 \\  a+1 = 0  \end{cases}   \\  \\ \begin{cases} b = 1 \\ a = -1   \end{cases}    \\  \\  \\ y = -x +1   \\  \\ y+x = 1 \\  \\ \boxed{4y+4x = 4}



\pmb{2^{o}} \\  \\ If \ A=(x_{1}, y_{1}); \ \ \ B = (x_{2},y_{2}) \\  \\ (x_{2} - x_{1}) (y - y_{1})=(y_{2}-y_{1})(x  -x_{1}) \\  \\ (1 - 0)(y - 1)=(0 - 1)(x - 0) \\  \\ y - 1 = -x \\  \\ y+x = 1 \\  \\ \boxed{4y+4x = 4}
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Maria studied the traffic trends in India. She found that the number of cars on the road increases by 10% each year. If there we
Delvig [45]

Answer:

Step-by-step explanation:

Let's first find the exponential function that models the situation in year one. The exponential standard form is

y=a(b)^x where a is the initial value and b is the growth/decay rate in decimal form. If it is growth it is added to 100% of the initial value; if it is decay it is taken away from 100% of the initial value. We are told that the number of cars in year one was 80 million, so

a = 80 (in millions)

If b is increasing by 10%, then we are adding that amount to the initial 100% we started with to give us 100% + 10% = 110% or, in decimal form, 1.1

The model for our situation is

y=80(1.1)^x where y is the number of cars after x years goes by. We want to find the difference between years 3 and 2, so we will use our model twice, replacing x with both a 2 and then a 3 and subtracting.

When x = 2:

y=80(1.1)^2 and

y = 80(1.21) so

y = 96.8 million cars

When x = 3:

y=80(1.1)^3 and

y = 80(1.331) so

y = 106.48 million cars

The difference between years 3 and 2 is

106.48 - 96.8 = 9.68 million cars

7 0
3 years ago
Evaluate the function for the given values of x.
galben [10]

Answer:

g(3) = 11

g(-3) = 16

g(-1) =3

Step-by-step explanation:

For g(3)

g(x)= x² +2

g(3) = 3² + 2

g(3) = 9+2

g(3) = 11

For g(-3)

g(x) = -3x + 7

g(-3) = -3(-3) + 7

g(-3) = 9 + 7

g(-3) = 16

For g(-1)

g(x) = x² + 2

g(-1) = (-1)² +2

g(-1) = 1+2

g(-1) =3

6 0
3 years ago
Pattern A follows the rule "add 5" and Pattern B follows the rule "subtract 2."
xxTIMURxx [149]

Answer:

(25,12)

Step-by-step explanation:

pattern A=20+5

=25

pattern B=14-2

=12

25,12

4 0
2 years ago
Factor the expression completely.<br> x^4y^4 – x^3y
enyata [817]

Answer:

x^3 y(xy^3−1)

7 0
2 years ago
Read 2 more answers
A shipment of 50,000 transistors arrives at a manufacturing plant. The quality control engineer at the plant obtains a random sa
Aleks04 [339]

Step-by-step explanation:

remember, the number of possible combinations to pick m out of n elements is C(n, m) = n!/(m! × (n-m)!)

50,000 transistors.

4% are defective, that means 4/100 = 1/25 of the whole.

so, the probability for one picked transistor to be defective is 1/25.

and the probability for it to work properly is then 1-1/25 = 24/25.

now, 500 picks are done.

to accept the shipment, 9 or less of these 500 picks must be defective.

the probability is then the sum of the probabilities to get

0 defective = (24/25)⁵⁰⁰

1 defective = (24/25)⁴⁹⁹×1/25 × C(500, 1)

= 24⁴⁹⁹/25⁵⁰⁰ × 500

2 defective = (24/25)⁴⁹⁸×1/25² × C(500, 2)

= 24⁴⁹⁸/25⁵⁰⁰ × 250×499

3 defective = 24⁴⁹⁷/25⁵⁰⁰ × C(500, 3) =

= 24⁴⁹⁷/25⁵⁰⁰ × 250×499×166

...

9 defective = 24⁴⁹¹/25⁵⁰⁰ × C(500, 9) =

= 24⁴⁹¹/25⁵⁰⁰ × 500×499×498×497×496×495×494×493×492×491 /

9×8×7×6×5×4×3×2 =

= 24⁴⁹¹/25⁵⁰⁰ × 50×499×166×71×31×55×494×493×41×491

best to use Excel or another form of spreadsheet to calculate all this and add it all up :

the probability that the engineer will accept the shipment is

0.004376634...

which makes sense, when you think about it, because 10 defect units in the 500 is only 2%. and since the whole shipment contains 4% defect units, it is highly unlikely that the random sample of 500 will pick so overwhelmingly the good pieces.

is the acceptance policy good ?

that completely depends on the circumstances.

what was the requirement about max. faulty rate in the first place ? if it was 2%, then the engineer's approach is basically sound.

it then further depends what are the costs resulting from a faulty unit ? that depends again on when the defect is usually found (still in manufacturing, or already out there at the customer site, or somewhere in between) and how critical the product containing such transistors is. e.g. recalls for products are extremely costly, while simply sorting the bad transistors out during the manufacturing process can be rather cheap. if there is a reliable and quick process to do so.

so, depending on repair, outage and even penalty costs it might be even advisable to have a harder limit during the sample test.

in other words - it depends on experience and the found distribution/probability curve, standard deviation, costs involved and other factors to define the best criteria for the sample test.

3 0
2 years ago
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