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love history [14]
2 years ago
14

predict the percentage of high school students who​ smoke, assuming that 25 25​% of adults in the region smoke. Use 25 25​, not

0. 25 25.
Mathematics
1 answer:
Vinvika [58]2 years ago
3 0
In most countries, students are not allowed to smoke, so this would be downright illegal. Also, students are educated about the detrimental effects of drugs so this would reduce it further. For the statistics, it would probably be less than 1%, under the circumstances. Please give brainliest, it would be much appreciated
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Compare 6.23 * 10^14 and 8.912 * 10^12
Masteriza [31]

Answer:

623,000,000,000,000 > 8,912,000,000,000

Step-by-step explanation:

8.912·10^12

The exponent is 12, making it 10 to the power of 12. As the exponent is positive, the solution is a number greater than the origin or base number. To find our answer, we move the decimal to the right 12 time(s):

8.912 -> 8,912,000,000,000

6.23·10^14

The exponent is 14, making it 10 to the power of 14. As the exponent is positive, the solution is a number greater than the origin or base number. To find our answer, we move the decimal to the right 14 time(s):

6.23 -> 623,000,000,000,000

6 0
3 years ago
URGENT PLEASE I AM TIMED URGENT SOLVE THE PROBLEM AND PLEASE, PLEASE EXPLAIN HOW YOU GOT YOUR ANSWER, GIVE DETAIL'S I WILL MARK
Neko [114]

I'm pretty sure you just multiply the 1 1/4 by 3 because if 1 1/4 makes 10 muffins you need to make 3 more dozens so 1 1/4 times 3 equals 3.75

3 0
3 years ago
NO LINKS OR FILES!
Archy [21]

(a) If the particle's position (measured with some unit) at time <em>t</em> is given by <em>s(t)</em>, where

s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}

then the velocity at time <em>t</em>, <em>v(t)</em>, is given by the derivative of <em>s(t)</em>,

v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}

(b) The velocity after 3 seconds is

v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}

(c) The particle is at rest when its velocity is zero:

\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|

In interval notation, this happens for <em>t</em> in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

\displaystyle \int_0^8 |v(t)|\,\mathrm dt

By definition of absolute value, we have

|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)

In part (d), we've shown that <em>v(t)</em> > 0 when -√11 < <em>t</em> < √11, so we split up the integral at <em>t</em> = √11 as

\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt

and by the fundamental theorem of calculus, since we know <em>v(t)</em> is the derivative of <em>s(t)</em>, this reduces to

s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}

7 0
2 years ago
5.75 , 7.75 , 9.75 , 11.75 is a Non linear sequence ? is it true or false
Tomtit [17]

Answer:

false because each go up by two and I didnt even look it up

5 0
3 years ago
Read 2 more answers
Please help<br>are these functions or not?​
crimeas [40]

Answer:

1. Function

2. Not a function

3. Function

4. Not a function

Step-by-step explanation:

A function just means that for each input it outputs only 1 output. This output isn't necessarily unique. So for example if you're just given: f(2) = 3 and f(1) = 3, this is a function since each input only outputs 1 value, even though the output isn't unique, but if you're given: f(3) = 2, f(2) = 1, f(3) = 3. that's not a function since f(3) outputs both 2 and 3.

Anyways now that you hopefully understand this, let's look at each image.


Relation 1: (Function)

   So for each input (the domain) it's only pointing to one output (range), even if multiple input (the domain) are pointing to the same output (the range), it's still a function.

Relation 2: (Not a function)

   This is not a function, since if you look at the input 7 (the range), you'll see it outputs two things (range). It outputs -6 and -7. So this is not a function

Relation 3: (Function)

   This is a function since each x-value (input) has only one y-value (output). So it's a function

Relation 4: (Not a functio)

   This is not a function, since there are multiply coordinates with the same x-coordinate (input) and different y-coordinates (output). The x-coordinate 2 has the output b, y, and m, and since no value is given for these variables, it can be assumed they're different values, thus it's not a function.

5 0
1 year ago
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