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2 years ago

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In a survey examining eating habit 34 women over 40 years old and a 54 women 40 and under responded to the survey. What’s the ra

tio of women over 40 to women 40 and under? Reduce the ratio

1 answer:

Rina8888 [55]2 years ago

4 0

Answer:

17:27

Step-by-step explanation:

You would start with 34:54

this reduces to 17:27

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The first three terms of a sequence are given. Round to the nearest thousandth (if necessary).

BabaBlast [244]

Step-by-step explanation:

the introduction of a fraction tells us that we are dealing with multiplications, and therefore a geometric sequence (where every new term is created by multiplying the previous term by a constant factor, the ratio r).

I think your teacher made a mistake, or you made one when typing the question in here.

there is no factor r that creates

15×r = 9

and

9×r = 5/27

it would mean that

15 × r² = 5/27

r² = 5/27 / 15 = 5/27 × 1/15 = 5/405 = 1/81

r = 1/9

but 15 × 1/9 = 5 × 1/3 = 5/3 is NOT 9

and 9 × 1/9 = 9/9 = 1 is NOT 5/27

so, this can't be right.

on the other hand

15 × r = 9

r = 9/15 = 3/5

and then

9 × 3/5 = 27/5

so, either the sequence should have been

15, 5/3, 5/27

or (and I suspect this to be true)

15, 9, 27/5

under that assumption we have

s1 = 15

r = 3/5

sn = sn-1 × r = s1 × r^(n-1) = 15 × (3/5)^(n-1)

s10 = 15 × (3/5)⁹ = 15 × 19683/1953125 =

= 3 × 19683/390625 = 59049/390625 =

= 0.15116544 ≈ 0.151

6 0

2 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

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

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2 years ago

Please help i will give metal.

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It would be either A or D because it says that john has 3 rows and does not say of how many stickers...but Raj have 36 stickers and Tajika has 36 stickers and Sue Lee has 40 so it would either A or D

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3 years ago

0.0432 in scientific notation

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The answer in scientific notation is: 4.32 x 10^-2

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3 years ago

Write two expressions that are equivalent to 2(x+4)

kotegsom [21]

(x+4)+(x+4)

2(x+2(2)) the (2) would be the exponent for 2 in the second one.

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3 years ago