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3 years ago

Body fat percentage was found to be 16.5% and the person weighs 124 pounds what part in pounds of the body weight is fat

Mathematics

1 answer:

puteri [66]3 years ago

7 0

Answer:

20.46 pounds

Step-by-step explanation:

Since 16.5% of their body is fat you need to multiply the weight by that percentage
you can do this by multiplying 124 by .165 which equals 20.46 pounds

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100 POINTS!!!!!!!!!!!!!!!!

mihalych1998 [28]

The change, from the predicted data to the actual data, in the average number of downloads of the application for Company A from the day the application was launched to 4 days after the application was launched would decrease by approximately 244 downloads per day.

The change, from the predicted data to the actual data, in the average number of downloads of the application for Company B from the day the application was launched to 4 days after the application was launched would increase by approximately 174 downloads per day.

Based on this information, Company B made a more accurate prediction of the average number of downloads of the application per day.

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2 years ago

A small box of cereal weighs k grams. A jumbo box of cereal weighs 5 times as much.

mariarad [96]

Answer:

k=m/5

Step-by-step explanation:

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2 years ago

A student wrote a number pattern such that each number is 3 more that 4 times the previous number. What is the sixth term in the

Nesterboy [21]

Start with 1.
1st number = 1
2nd number = 1 * 4 + 3 = 7
3rd number = 7 * 4 + 3 = 31
4th = 31 * 4 + 3 = 127
5th = 127 * 4 + 3 = 511
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3 years ago

Read 2 more answers

Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut

frozen [14]

Answer:

a) $\sqrt{144-288t+208t^2}$ b.) -12knots, 8 knots c) No e) $4\sqrt{13}$

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

We know that at time t , the ship A has moved $12\dot t (n.m)$ and ship B has moved $8\dot t (n.m)$ . We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

$\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}$

We want to find $\frac{ds}{dt}$ for t=0 and t=1

$\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots$

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = $s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0$

Since function $f(x)=199-288x+208x^2$ is quadratic, concave up and has no real roots, we know that $199-288x+208x^2>0$ for every t. So, the ships haven't seen each other.

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

Function ds/dt has a horizontal asympote in the first quadrant if

$\lim_{t \to \infty} \frac{ds}{dt}$

So, lets check this limit:

$\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}$

Notice that:

$4\sqrt{13}=\sqrt{12^2+5^2}$ =√(speed of ship A² + speed of ship B²)

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3 years ago

Please please help i have this midterm due tomorrow

jek_recluse [69]

You should know that cos2x =1-2sin^2x, so you substitute and take everything to one side. After that I took out a common factor of sinx which simplified my equation and gave me my first solution x=0. I go and find the relative acute angle in my second solution 2sinx+1=0, and as the value of raw is negative you have to obtain two solutions from where sin is negative. And you are given all you solutions. Hopefully this helps, if you don’t understand what I am saying just ask me anything. :)

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2 years ago

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