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1 month ago

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Sarah can bicycle a loop around the north part of Lake Washington in 2 hours and 30 minutes. If she could increase her average s

peed by 1 km/hr, it would reduce her time around the loop by 7 minutes. How many kilometers long is the loop?

1 answer:

Leona [12.6K]1 month ago

7 0

Response:

The loop's length is roughly 49.58 km.

Step-by-step details:

Given:

Duration to cycle the loop = 2 hours and 30 minutes

It is known that;

60 minutes = 1 hour

30 minutes = 0.5 hour

therefore, 2 hours and 30 minutes = $2+0.5 = 2.5\ hrs$

∴ Time to cycle the loop = 2.5 hours

Let 's' denote the cycling speed.

Denote the total loop distance as 'd'

From our understanding;

Distance equals speed times time.

Representing this as an equation we have;

$d =2.5s \ \ \ \ equation \ 1$

Given:

If her speed increased by 1 km/hr, the time taken around the loop would decrease by 7 minutes.

As a result, we can say;

Speed = $s+1$

The time then will be shortened by 7 minutes.

7 minutes = 0.12 hours

Consequently, time = 2.5 - 0.12 = 2.38

Again, Distance is calculated as speed times time.

This can be expressed with equations;

$d =2.38(s+1) \ \ \ \ equation \ 2$

The distances are equivalent in both scenarios, thus we will determine speed by equating the equations.

$2.5s = 2.38(s+1)\\\\2.5s=2.38s+2.38\\\\2.5s-2.38s =2.38\\\\0.12s= 2.38\\\\s =\frac{2.38}{0.12} \approx 19.83\ km/hr$

So, Speed = 19.83 km/hr

Loop length (d) = $19.83\times 2.5 = 49.58 \ km$

Therefore, the loop length is approximately 49.58 km.

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Response: C. 1.942

Detailed explanation:

This analysis involves comparing two population proportions. The rate of absenteeism among employees in plant A and plant B is denoted by p1 and p2 respectively.

From the provided data,

p1 = 0.15

p2 = 0.12

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n2 = 1200

To find the z score, we start by calculating the pooled proportion.

The pooled proportion, pc is

pc = (x1 + x2)/(n1 + n2)

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1 - pc = 1 - 0.132 = 0.868

The z score formula is

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21 day ago

Gabi wants to drive to and from the airport. She finds two companies near her that offer short-term car rental service at differ

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Answer:

The cost difference per mile between the two companies is $0.12.

Step-by-step explanation:

Gabi formulates the equation $0.22m+7.20=0.1m+8.40$ to determine after how many miles, denoted as m, the charges of both companies will be equal.

The first company levies $c_1=0.22m+7.20$ for m miles traveled.

The second company's charge for the same m miles is $c_2=0.1m+8.40$ .

In these equations, the figures 7.20 and 8.40 signify the initial fees the companies impose.

The values 0.22 and 0.1 represent the respective costs per mile.

As such, the disparity in per-mile charges amounts to $0.22-0.1=0.12$ .

An alternative method to tackle this problem is by calculating the per-mile rate for each company:

1. Cost per mile for the first company

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2. Cost per mile for the second company

$c_2(0)=0.1\cdot 0+8.40=8.40\\ \\c_2(1)=0.1\cdot 1+8.40=8.50\\ \\c_2(1)-c_2(0)=8.50-8.40=0.1$

3. The difference:

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1 month ago

Kevin is buying water for his camping trip. He knows he needs at least 20 gallons of water for the trip. He already has five and

PIT_PIT [12445]

Response:

$5.5+0.25x\geq 20$

Step-by-step breakdown:

Kevin has already gathered five and a half gallons of water for his trip

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Let x represent the number of 32-fluid ounce (quarter-gallon) containers needed to collect at least 20 gallons of water for the trip.

One container holds 0.25 gallons of water

Therefore, x containers hold 0.25x gallons of water

Thus, Kevin's total gallons of water = $5.5+0.25x$

Since it is given that he needs at least 20 gallons of water for the trip.

Hence, $5.5+0.25x\geq 20$

Thus, the algebraic inequality representing this scenario is $5.5+0.25x\geq 20$

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27 days ago

Read 2 more answers

Find the point on the circle x^2+y^2 = 16900 which is closest to the interior point (30,40)

Leona [12618]

Response-

(78,104) represents the point closest to the interior.

Explanation-

The equation defining the circle,

$\Rightarrow x^2+y^2 = 16900$

$\Rightarrow y^2 = 16900-x^2$

$\Rightarrow y = \sqrt{16900-x^2}$

Since the point lies on the circle, its coordinates must be,

$(x,\sqrt{16900-x^2})$

The distance "d" from the point to (30,40) can be calculated as,

$=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$=\sqrt{(x-30)^2+(\sqrt{16900-x^2}-40)^2}$

$=\sqrt{x^2+900-60x+16900-x^2+1600-80\sqrt{16900-x^2}}$

$=\sqrt{9400-60x-80\sqrt{16900-x^2}}$

Next, we need to determine the value of x for which d is minimized. The minimum distance occurs when $9400-60x-80\sqrt{16900-x^2}$ is at its lowest value.

Let’s set up the equation,

$\Rightarrow f(x)=9400-60x-80\sqrt{16900-x^2}$

$\Rightarrow f'(x)=-60+80\dfrac{x}{\sqrt{16900-x^2}}$

$\Rightarrow f''(x)=\dfrac{1352000}{\left(16900-x^2\right)\sqrt{16900-x^2}}$

We find the critical points,

$\Rightarrow f'(x)=0$

$\Rightarrow-60+80\dfrac{x}{\sqrt{16900-x^2}}=0$

$\Rightarrow 80\dfrac{x}{\sqrt{16900-x^2}}=60$

$\Rightarrow 80x=60\sqrt{16900-x^2}$

$\Rightarrow 80^2x^2=60^2(16900-x^2)$

$\Rightarrow 6400x^2=3600(16900-x^2)$

$\Rightarrow \dfrac{16}{9}x^2=16900-x^2$

$\Rightarrow \dfrac{25}{9}x^2=16900$

$\Rightarrow x=\sqrt{\dfrac{16900\times 9}{25}}=78$

$\Rightarrow x=78$

Then,

$\Rightarrow f''(78)=\dfrac{1352000}{\left(16900-78^2\right)\sqrt{16900-78^2}}=\dfrac{125}{104}=1.2$

Since f''(x) is positive, the function f(x) achieves its minimum at x=78

When x is set to 78, the corresponding y value will be

$\Rightarrow y = \sqrt{16900-x^2}=\sqrt{16900-78^2}=104$

This leads us to conclude that the closest point is (78,104)

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1 month ago

The equation of the tangent plane to the ellipsoid x2/a2 + y2/b2 + z2/c2 = 1 at the point (x0, y0, z0) can be written as xx0 a2

PIT_PIT [12445]

Answer:

The tangent plane equation for the hyperboloid

$\frac{xx_0}{a^2}+\frac{yy_0}{b^2}-\frac{zz_0}{c^2}=1$ .

Step-by-step explanation:

We have

The ellipsoid's equation is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$

The equation for the tangent plane at the point $\left(x_0,y_0,z_0\right)$

$\frac{xx_0}{a^2}+\frac{yy_0}{b^2}+\frac{zz_0}{c^2}=1$ (Given)

The hyperboloid's equation is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$

F(x,y,z)= $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}[c^2}$

$F_x=\frac{2x}{a^2},F_y=\frac{2y}{b^2},F_z=-\frac{2z}{c^2}$

$(F_x,F_y,F_z)(x_0,y_0,z_0)=\left(\frac{2x_0}{a^2},\frac{2y_0}{b^2},-\frac{2z_0}{c^2}\right)$

The tangent plane equation at point $\left(x_0,y_0,z_0\right)$

$\frac{2x_0}{a^2}(x-x_0)+\frac{2y_0}{b^2}(y-y_0)-\farc{2z_0}{c^2}(z-z_0)=0$

The tangent plane equation for the hyperboloid is

$\frac{2xx_0}{a^2}+\frac{2yy_0}{b^2}-\frac{2zz_0}{c^2}-2\left(\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}-\frac{z_0^2}{c^2}\right)=0$

The tangent plane equation

$2\left(\frac{xx_0}{a^2}+\frac{yy_0}{b^2}-\frac{zz_0}{c^2}\right)=2$

Hence, the required tangent plane equation for the hyperboloid is

$\frac{xx_0}{a^2}+\frac{yy_0}{b^2}-\frac{zz_0}{c^2}=0$

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1 month ago