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dusya [7]
3 years ago
5

What would be the appropriate units and intervals to use along the x- and y-axes

Mathematics
1 answer:
Vika [28.1K]3 years ago
3 0

X axes are supposed to be positive and the y is negative.

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The opposite of the fraction one third
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The opposite of the fraction one third would be negative one third. To be opposite, they must have differing signs. One number should be positive and the other number should be negative. It is different from reciprocal. To be a reciprocal, <span>one number should be the flipped fraction, or upside down version, of the other number.</span>
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A fair coin and a biased coin where the probability of obtaining a head is twice the probability of obtaining a tail. if both ar
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Answer:

See below in bold.

Step-by-step explanation:

For the fair coin Prob(head) = 1/2 and Prob(Tail) = 1/2.

For the biased coin it is   Prob(head) = 2/3 and Prob(Tail) = 1/3.

a) Prob(2 heads) = 1/2 * 2/3 = 1/3.

b) Prob(2 tails) = 1/2 * 1/3 = 1/6.

c) Prob(1 head ) = Prob(H T or T H) = 1/2 * 1/3  + 1/2 * 2/3) = 1/6+1/3 = 1/2.

d) Prob (at least one head) = prob (HH or TH or HT) =  1/3 + 1/2 =<em> </em>5/6.

4 0
3 years ago
PLS HELP ME IM ON TIME4LEARNING
alukav5142 [94]

Answer:

13

Step-by-step explanation:

12*12=144

5*5=25

\sqrt{169}  = 13

7 0
2 years ago
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Ann [662]

Proportional

Both equal 6/15

6 0
3 years ago
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Help please i don’t know how to do it
Norma-Jean [14]

Answer:

Step-by-step explanation:

\frac{dy}{dx}=x^{2}(y-1)\\\frac{1}{y-1} \text{ } dy=x^{2} \text{ } dx\\\int \frac{1}{y-1} \text{ } dy=\int x^{2} \text{ } dx\\\ln|y-1|=\frac{x^{3}}{3}+C\\

From the initial condition,

\ln|3-1|=\frac{0^{3}}{3}+C\\\ln 2=C

So we have that \ln |y-1|=\frac{x^{3}}{3}+\ln 2\\e^{\frac{x^{3}}{3}+\ln 2}=y-1\\2e^{\frac{x^{3}}{3}}=y-1\\y=2e^{\frac{x^{3}}{3}}+1

4 0
3 years ago
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