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e-lub [12.9K]
3 years ago
14

In a survey of a community, it was found that 85% of the people like winter season and 65% like summer season. If none of them d

id not like both seasons
i) what percent like both the seasons
​
Mathematics
1 answer:
KengaRu [80]3 years ago
6 0

Answer:

50%

Step-by-step explanation:

Let :

Winter = W

Summer = S

P(W) = 0.85

P(S) = 0.65

Recall:

P(W u S) = p(W) + p(S) - p(W n S)

Since, none of them did not like both seasons, P(W u S) = 1

Hence,

1 = 0.85 + 0.65 - p(both)

p(both) = 0.85 + 0.65 - 1

p(both) = 1.50 - 1

p(both) = 0.5

Hence percentage who like both = 0.5 * 100% = 50%

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What is the coefficient of (3y^2 + 9)5
PtichkaEL [24]

The coefficient of (3y² + 9)5 is <u>15</u>.

A polynomial is of the form a₀xⁿ + a₁xⁿ⁻¹ + a₂xⁿ⁻² + ... + aₙ₋₁x + aₙ.

Here, x is the variable, aₙ is the constant term, and a₀, a₁, a₂, ..., and aₙ₋₁, are the coefficients.

a₀ is the leading coefficient.

In the question, we are asked to identify the coefficient of (3y² + 9)5.

First, we expand the given expression:

(3y² + 9)5

= 15y² + 45.

Comparing this to the standard form of a polynomial, a₀xⁿ + a₁xⁿ⁻¹ + a₂xⁿ⁻² + ... + aₙ₋₁x + aₙ, we can say that y is the variable, 15 is the coefficient, and 45 is the constant term.

Thus, the coefficient of (3y² + 9)5 is <u>15</u>.

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6 0
2 years ago
Seven less than the product of a number n and
SVETLANKA909090 [29]

Answer:

(n-7) ≤ 95

Step-by-step explanation:

We need to write the expression for the given statement.

Seven less than the product of a number n and  1 means (n-7).

It is no more than 95. No more than 95 means less than or equal to i.e.

(n-7) ≤ 95

Hence, the above inequality shows the given statement.

5 0
3 years ago
An asset that costs $14,400 and has accumulated depreciation of $8,000 is sold for $5,600. What amount of gain or loss will be r
Advocard [28]

Answer:

There will be an $800 loss

Step-by-step explanation:

The initial cost of the asset is $14,400

When an asset depreciates, it loses its potential value over time. In this case, the total depreciation of our asset is given as $8,000.

The Current value of the asset = Initial Value - Accumulated Depreciation

Current value = $14,400 - $8,000 = $6,400

Since the asset is sold for $5,600, it is clear that this is a loss, as the selling price of $5,600 is less than the actual price of $6,400. So,

Loss = Actual (Cost) Price - Selling Price = $6,400 - $5,600 = $800

6 0
3 years ago
Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of
kifflom [539]

Looks like we have

\vec F(x,y,z)=z^2x\,\vec\imath+\left(\dfrac{y^3}3+\sin z\right)\,\vec\jmath+(x^2z+y^2)\,\vec k

which has divergence

\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(z^2x)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial z}=z^2+y^2+x^2

By the divergence theorem, the integral of \vec F across S is equal to the integral of \nabla\cdot\vec F over R, where R is the region enclosed by S. Of course, S is not a closed surface, but we can make it so by closing off the hemisphere S by attaching it to the disk x^2+y^2\le1 (call it D) so that R has boundary S\cup D.

Then by the divergence theorem,

\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(x^2+y^2+z^2)\,\mathrm dV

Compute the integral in spherical coordinates, setting

\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi

so that the integral is

\displaystyle\iiint_R(x^2+y^2+z^2)\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^1\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{2\pi}5

The integral of \vec F across S\cup D is equal to the integral of \vec F across S plus the integral across D (without outward orientation, so that

\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\iint_D\vec F\cdot\mathrm d\vec S

Parameterize D by

\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath

with 0\le u\le1 and 0\le v\le2\pi. Take the normal vector to D to be

\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=-u\,\vec k

Then we have

\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^1\left(\frac{u^3}3\sin^3v\,\vec\jmath+u^2\sin^2v\,\vec k\right)\times(-u\,\vec k)\,\mathrm du\,\mathrm dv

=\displaystyle-\int_0^{2\pi}\int_0^1u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac\pi4

Finally,

\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\left(-\frac\pi4\right)=\boxed{\frac{13\pi}{20}}

6 0
4 years ago
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