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2 years ago

10000 people were surveyed and 60% liked pizza. What is the absolute value equation that describes the interval?

1 answer:

6 0

The statement says, "10000 people were surveyed and 60% like pizza." This means that

Number of people like pizza = 10000 * 0.60
Number of people like pizza = 6000

The question asked for the absolute value equation. This means that
|10000 * 0.60| is the absolute value equation

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PLEASE HELP!!!!! AND TAHNK YOU

Anarel [89]

Answer:

a= 68% times 250, there are 170 students not in Advanced

Step-by-step explanation:

If 32% are in Adv math, 68% aren't, to solve a percentage, I did

.68 times 250

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2 years ago

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Statistics urgent<br> Use n=6 and p=0.85 to complete

Bas_tet [7]

Answer:

I cant see anything on that picture. Wish I could help bruv. I wish I could help. It's a shame innit.

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2 years 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

<img src="https://tex.z-dn.net/?f=-2%5Csqrt%7B50x%5E5%7D%20%2B5%5Csqrt%7B18x%5E5%7D" id="TexFormula1" title="-2\sqrt{50x^5} +5\s

MaRussiya [10]

Answer:

234.98

Step-by-step explanation:

in class notes

3 0

2 years ago

denpristay [2]

Answer:

Actual height of the building is 670 m

Step-by-step explanation:

We have been the actual height of a Ferris wheel which is 134 m. Along with it we are also given the scaled down heights of a Ferris wheel and a building. Using this data we have to find the actual height of a building.

Actual height of Ferris wheel = 134 m.

Height of Ferris wheel using the scale = 0.65 cm

Height of building after using the scale - 3.25 cm

Let, the actual height of building be x meters.

Since, the same scale is being used, the ratio of actual heights to that of scaled down heights must be the same for both objects.

i.e.

Ratio of Actual Height to scale height of Ferris Wheel = Ratio of Actual Height to scale height of the building

Using the values, we get:

$134:0.65 =x:3.25\\\\\frac{134}{0.65}=\frac{x}{3.25}\\\\ \frac{134}{0.65} \times 3.25 = x\\\\ x=670$

The units of actual heights are in meters, so this means the actual height of the building is 670 m

5 0

2 years ago