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4 years ago

7/8 students own a bike . 72 is the difference between the ones that own a bike and ones that do not. how many were surveyed?

Mathematics

2 answers:

uranmaximum [27]4 years ago

7 0

The answer is 96 students were surveyed

DerKrebs [107]4 years ago

6 0

X + y is the number of students who were surveyed. x is the number who own a bike. y is the number who don't. the ratio of the ones who own a bike and the ones who were surveyed is x / (x + y) that ratio is 7/8. you get x / (x + y) = 7/8 multiply both sides of this equation by (x + y) to get x = (x + y) * 7/8 simplify to get x = 7/8 * x + 7/8 * y subtract 7/8 * x from both sides of this equation to get: 1/8 * x = 7/8 * y solve for x to get x = 7 * y this means that the number of students who own a bike is 7 times the number of students who don't. divide both sides of that equation by y to get x/y = 7/1. that means that the ratio of students who own a bike to the number of students who don't is equal to 7/1. the problem states that the difference between the number who own a bike and the number who don't is 72. this means that x - y = 72 from the equation of x = 7 * y, replace x with 7 * y in the equation of x - y = 72 to get: 7 * y - y = 72 simplify to get 6 * y = 72 divide both sides of that equation by 6 to get y = 72/6 = 12 this means that the number of students who do not own a bike is 12. since x = 7 * y, then x = 7 * 12 = 84. this means that the number of students who own a bike is 84. since the number of students is x + y, this means that the number of students who were surveyed is 84 + 12 = 96. let's see if this is right. 96 students were surveyed. 84 own a bike and 12 don't. the difference between the ones who own a bike and the ones who don't is 84 - 12 = 72. the ratio of the ones who own a bike and the number surveyed is 84/96 = 7/8. solution is 96 students were surveyed.

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Find the exact area of the surface obtained by rotating the curve about the x-axis. y = 1 + ex , 0 ≤ x ≤ 9

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$\displaystyle2\pi\int_0^9(1+e^x)\sqrt{1+e^{2x}}\,\mathrm dx$

since $y=1+e^x\implies y'=e^x$ . To compute the integral, first let

$u=e^x\implies x=\ln u$

so that $\mathrm dx=\frac{\mathrm du}u$ , and the integral becomes

$\displaystyle2\pi\int_1^{e^9}\frac{(1+u)\sqrt{1+u^2}}u\,\mathrm du$

$=\displaystyle2\pi\int_1^{e^9}\left(\frac{\sqrt{1+u^2}}u+\sqrt{1+u^2}\right)\,\mathrm du$

Next, let

$u=\tan t\implies t=\tan^{-1}u$

so that $\mathrm du=\sec^2t\,\mathrm dt$ . Then

$1+u^2=1+\tan^2t=\sec^2t\implies\sqrt{1+u^2}=\sec t$

so the integral becomes

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\left(\frac{\sec t}{\tan t}+\sec t\right)\sec^2t\,\mathrm dt$

$=\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\left(\frac{\sec^3t}{\tan t}+\sec^3 t\right)\,\mathrm dt$

Rewrite the integrand with

$\dfrac{\sec^3t}{\tan t}=\dfrac{\sec t\tan t\sec^2t}{\sec^2t-1}$

so that integrating the first term boils down to

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\frac{\sec t\tan t\sec^2t}{\sec^2t-1}\,\mathrm dt=2\pi\int_{\sqrt2}^{\sqrt{1+e^{18}}}\frac{s^2}{s^2-1}\,\mathrm ds$

where we substitute $s=\sec t\implies\mathrm ds=\sec t\tan t\,\mathrm dt$ . Since

$\dfrac{s^2}{s^2-1}=1+\dfrac12\left(\dfrac1{s-1}-\dfrac1{s+1}\right)$

the first term in this integral contributes

$\displaystyle2\pi\int_{\sqrt2}^{\sqrt{1+e^{18}}}\left(1+\frac12\left(\frac1{s-1}-\frac1{s+1}\right)\right)\,\mathrm ds=2\pi\left(s+\frac12\ln\left|\frac{s-1}{s+1}\right|\right)\bigg|_{\sqrt2}^{\sqrt{1+e^{18}}}$

$=2\pi\sqrt{1+e^{18}}+\pi\ln\dfrac{\sqrt{1+e^{18}}-1}{1+\sqrt{1+e^{18}}}$

The second term of the integral contributes

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\sec^3t\,\mathrm dt$

The antiderivative of $\sec^3t$ is well-known (enough that I won't derive it here myself):

$\displaystyle\int\sec^3t\,\mathrm dt=\frac12\sec t\tan t+\frac12\ln|\sec t+\tan t|+C$

so this latter integral's contribution is

$\pi\left(\sec t\tan t+\ln|\sec t+\tan t|\right)\bigg|_{\pi/4}^{\tan^{-1}(e^9)}=\pi\left(e^9\sqrt{1+e^{18}}+\ln(e^9+\sqrt{1+e^{18}})-\sqrt2-\ln(1+\sqrt2)\right)$

Then the surface area is

$2\pi\sqrt{1+e^{18}}+\pi\ln\dfrac{\sqrt{1+e^{18}}-1}{1+\sqrt{1+e^{18}}}+\pi\left(e^9\sqrt{1+e^{18}}+\ln(e^9+\sqrt{1+e^{18}})-\sqrt2-\ln(1+\sqrt2)\right)$

$=\boxed{\left((2+e^9)\sqrt{1+e^{18}}-\sqrt2+\ln\dfrac{(e^9+\sqrt{1+e^{18}})(\sqrt{1+e^{18}}-1)}{(1+\sqrt2)(1+\sqrt{1+e^{18}})}\right)\pi}$

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4 years ago