Compare 9/15 and 10/15

Mathematics

1 answer:

choli [55]2 years ago

6 0

Answer: 9/15 < 10/15

Step-by-step explanation: 9/15 < 10/15

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Step-by-step explanation:

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Simplify: –3(y + 2)2 – 5 + 6yWhat is the simplified product in standard form?

sweet-ann [11.9K]

Answer by Mimiwhatsup:

$\mathrm{Multiply\:the\:numbers:}\:3\cdot \:2=6\\=-6\left(y+2\right)-5+6y\\-6\left(y+2\right)\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac\\a=-6,\:b=y,\:c=2\\=-6y+\left(-6\right)\cdot \:2\\Apply\:minus-plus\:rules\\+\left(-a\right)=-a\\=-6y-6\cdot \:2\\\mathrm{Multiply\:the\:numbers:}\:6\cdot \:2=12\\=-6y-12\\=-6y-12-5+6y\\\mathrm{Simplify}\:-6y-12-5+6y\\Group\:like\:terms\\=-6y+6y-12-5\\\mathrm{Add\:similar\:elements:}\:-6y+6y=0\\=-12-5\\$

$\mathrm{Subtract\:the\:numbers:}\:-12-5=-17\\Answer is: =-17$

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3 years ago

Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species

kotegsom [21]

Answer:

$P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}$

Step-by-step explanation:

The logistic equation is the following one:

$P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}$

In which P(t) is the size of the population after t years, K is the carrying capacity of the population, r is the decimal growth rate of the population and P(0) is the initial population of the lake.

In this problem, we have that:

Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 2,000. This means that $P(0) = 80, K = 2000$ .