On the first season of "The Biggest Loser",

Mathematics

1 answer:

amid [387]2 years ago

4 0

Answer:

33%

Step-by-step explanation:

311-208=103 ponds lost

103/311=33%

You might be interested in

Frank thinks he will need at about 12 1/2 kilograms of sand for sandbox he is building. The sand he wants is 3$ per kilogram. Ab

vesna_86 [32]

12 * 3 = 36, 1/2 * 3 = 1 1/2
36 + 1 1/2 = $37.50 for about 13 kilograms of sand.

4 0

2 years ago

Four hundred people were asked whether gun laws should be more stringent. three hundred said "yes," and 100 said "no." the point

Anna11 [10]

To find relative frequency of no voters;

p(A)=favourable outcomes/total outcomes

p(A)=100/400

= 1/4 or 0.25

Therefore, the probability a person will respond with a no is 0.25.

Hope I helped :)

8 0

2 years ago

I need help with this can someone please help me I'm bad at fractions.

Artemon [7]

The answer is 29, I just didn’t on my head right now.

5 0

2 years ago

The population, P(t), of China, in billions, can be approximated by1 P(t)=1.394(1.006)t, where t is the number of years since th

vitfil [10]

Answer:

At the start of 2014, the population was growing at 8.34 million people per year.

At the start of 2015, the population was growing at 8.39 million people per year.

Step-by-step explanation:

To find how fast was the population growing at the start of 2014 and at the start of 2015 we need to take the derivative of the function with respect to t.

The derivative shows by how much the function (the population, in this case) is changing when the variable you're deriving with respect to (time) increases one unit (one year).

We know that the population, P(t), of China, in billions, can be approximated by $P(t)=1.394(1.006)^t$

To find the derivative you need to:

$\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=\\\\\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'\\\\1.394\frac{d}{dt}\left(1.006^t\right)\\\\\mathrm{Apply\:the\:derivative\:exponent\:rule}:\quad \frac{d}{dx}\left(a^x\right)=a^x\ln \left(a\right)\\\\1.394\cdot \:1.006^t\ln \left(1.006\right)\\\\\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=(1.394\cdot \ln \left(1.006\right))\cdot 1.006^t$

To find the population growing at the start of 2014 we say t = 0

$P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(0)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^0\\P(0)' = 0.00833901 \:Billion/year$

To find the population growing at the start of 2015 we say t = 1

$P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(1)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^1\\P(1)' = 0.00838904 \:Billion/year$