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3 years ago

How do you divide 2\3 divided by 7\8 using kcf answer

Mathematics

1 answer:

xz_007 [3.2K]3 years ago

7 0

Answer: $\frac{16}{21}$

Step-by-Step Explanation:

KCFmeans: Keep, Change, Flip

Keep: the first fraction

Change: the sign from division to multiplication

Flip: the second fraction

$\frac{2}{3}$ ÷ $\frac{7}{8}$

= $\frac{2}{3} * \frac{8}{7}$ KCF is applied here!

= $\frac{2(8)}{3(7)}$

= $\frac{16}{21}$

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3 years ago

As you were organizing your closet you counted 20 shirts and 12 pairs of pants what was the pants to shirt ratio?

NISA [10]

Answer: 5 : 3

Step-by-step explanation:

The ratio of shirts to pants can be shown by putting the number if shirts and pants in the following manner:

Number of shirts : Number of pants

20 : 12

Then you have to take this to the lowest term. 4 is a factor of both numbers so when divided by 4 we get:

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3 years ago

Which of the following random variables should be considered continuous? Select one: a. The number of cars owned by a randomly c

kolbaska11 [484]

The Random Variable that should be considered continuous is "The time it takes for a randomly chosen athlete to run 200 meters" , the correct option is (c) .

In the question ,

four random variables are given , we have to select the random variable that is continuous .

A Continuous Random Variable is a random variable X , which can take any value in its domain or in an interval or the union of intervals on the real line is called Continuous random variable .

For Example : weight of middle aged people lying between 40 Kg and 150 Kg .

From the given options ,

the option (c) , the time taken for a randomly chosen athlete to run 200 meters is a interval of time ,

So ,it is can be called as Continuous Random Variable .

Therefore , the option that represents Continuous Random Variable is option(c) .

The given question is incomplete , the complete question is

Which of the following random variables should be considered continuous ? Select one:

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1 year ago

A tank contains 180 gallons of water and 15 oz of salt. water containing a salt concentration of 17(1+15sint) oz/gal flows into

Stels [109]

Let A(t) denote the amount of salt (in ounces, oz) in the tank at time t (in minutes, min).

Salt flows in at a rate of

$\dfrac{dA}{dt}_{\rm in} = \left(17 (1 + 15 \sin(t)) \dfrac{\rm oz}{\rm gal}\right) \left(8\dfrac{\rm gal}{\rm min}\right) = 136 (1 + 15 \sin(t)) \dfrac{\rm oz}{\min}$

and flows out at a rate of

$\dfrac{dA}{dt}_{\rm out} = \left(\dfrac{A(t) \, \mathrm{oz}}{180 \,\mathrm{gal} + \left(8\frac{\rm gal}{\rm min} - 8\frac{\rm gal}{\rm min}\right) (t \, \mathrm{min})}\right) \left(8 \dfrac{\rm gal}{\rm min}\right) = \dfrac{A(t)}{180} \dfrac{\rm oz}{\rm min}$

so that the net rate of change in the amount of salt in the tank is given by the linear differential equation

$\dfrac{dA}{dt} = \dfrac{dA}{dt}_{\rm in} - \dfrac{dA}{dt}_{\rm out} \iff \dfrac{dA}{dt} + \dfrac{A(t)}{180} = 136 (1 + 15 \sin(t))$

Multiply both sides by the integrating factor, $e^{t/180}$ , and rewrite the left side as the derivative of a product.

$e^{t/180} \dfrac{dA}{dt} + e^{t/180} \dfrac{A(t)}{180} = 136 e^{t/180} (1 + 15 \sin(t))$

$\dfrac d{dt}\left[e^{t/180} A(t)\right] = 136 e^{t/180} (1 + 15 \sin(t))$

Integrate both sides with respect to t (integrate the right side by parts):

$\displaystyle \int \frac d{dt}\left[e^{t/180} A(t)\right] \, dt = 136 \int e^{t/180} (1 + 15 \sin(t)) \, dt$

$\displaystyle e^{t/180} A(t) = \left(24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t)\right) e^{t/180} + C$

Solve for A(t) :

$\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) + C e^{-t/180}$

The tank starts with A(0) = 15 oz of salt; use this to solve for the constant C.

$\displaystyle 15 = 24,480 - \frac{66,096,000}{32,401} + C \implies C = -\dfrac{726,594,465}{32,401}$

So,

$\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) - \frac{726,594,465}{32,401} e^{-t/180}$

Recall the angle-sum identity for cosine:

$R \cos(x-\theta) = R \cos(\theta) \cos(x) + R \sin(\theta) \sin(x)$

so that we can condense the trigonometric terms in A(t). Solve for R and θ :

$R \cos(\theta) = -\dfrac{66,096,000}{32,401}$

$R \sin(\theta) = \dfrac{367,200}{32,401}$

Recall the Pythagorean identity and definition of tangent,

$\cos^2(x) + \sin^2(x) = 1$

$\tan(x) = \dfrac{\sin(x)}{\cos(x)}$

Then

$R^2 \cos^2(\theta) + R^2 \sin^2(\theta) = R^2 = \dfrac{134,835,840,000}{32,401} \implies R = \dfrac{367,200}{\sqrt{32,401}}$

and

$\dfrac{R \sin(\theta)}{R \cos(\theta)} = \tan(\theta) = -\dfrac{367,200}{66,096,000} = -\dfrac1{180} \\\\ \implies \theta = -\tan^{-1}\left(\dfrac1{180}\right) = -\cot^{-1}(180)$

so we can rewrite A(t) as

$\displaystyle A(t) = 24,480 + \frac{367,200}{\sqrt{32,401}} \cos\left(t + \cot^{-1}(180)\right) - \frac{726,594,465}{32,401} e^{-t/180}$

As t goes to infinity, the exponential term will converge to zero. Meanwhile the cosine term will oscillate between -1 and 1, so that A(t) will oscillate about the constant level of 24,480 oz between the extreme values of

$24,480 - \dfrac{267,200}{\sqrt{32,401}} \approx 22,995.6 \,\mathrm{oz}$

and

$24,480 + \dfrac{267,200}{\sqrt{32,401}} \approx 25,964.4 \,\mathrm{oz}$

which is to say, with amplitude

$2 \times \dfrac{267,200}{\sqrt{32,401}} \approx \mathbf{2,968.84 \,oz}$

6 0

2 years ago

It's easy it's called Two-way tables please help

Vesnalui [34]

Answer: The filled out table is shown below in the attached image

===========================================================

Explanation:

The letters A through J are one way to fill out the table. Though there are other possible orders.

Start at letter A, which is from the fact we have 40 people total

Then move to letter B. This is the total number who like pasta (26)

We have C = A-B = 40-26 = 14 people who don't like pasta

--------------

Now move to cell D. This is the total who don't like netball, which is 21 people. There are E = A-D = 40-21 = 19 people who do like netball.

F = 4 is the number of people who don't like either thing (netball or pasta). So G = D-F = 21-4 = 17 people like pasta but don't like netball.

Of those people who like netball (E = 19), we have 9 who also like pasta. This is in cell H in the upper left corner. This means J = E-H = 19-9 = 10 people like netball but don't like pasta.

----------------

So we have these values

A = 40
B = 26
C = 14
D = 21
E = 19
F = 4
G = 17
H = 9
J = 10

7 0

3 years ago