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1 year ago

Jason Reads 126 pages in 6 hours. Joe reads 115 pages in 5 hours how many more pages per hour does Joe read?

1 answer:

4 0

Jason --> 126/6 = 21 pages per hour

joe --> 115/5 = 23 pages per hour

joe reads 2 more pages per hour

hope this helps :)

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You offer senior citizens a 20 percent discount on their meal prices served at your cafeteria. Assuming that an average of 150 s

scoray [572]

Given, there are an average of 150 citizens eat daile in the cafeteria.

The average price of their meals = $5.95.

The discount given for sinior citizens in their meals = 20%

So we have to find 20% of $5.95 first.

20% of $5.95 = $(5.95)(\frac{20}{100} )$

= $\frac{(5.95)(20)}{100}$

= $\frac{119}{100}$

= $1.19$

So, the amount of discount given to the senior citizens = $1.19 per head.

As there are 150 cinior citizens daily.

So total discount given daily = $ $(150)(1.19)$ =$ $178.5$

So we have got the required answer.

Option C is correct here.

3 0

2 years ago

Read 2 more answers

Suppose that a person's birthday is a uniformly random choice from the 365 days of a year (leap years are ignored), and one pers

maxonik [38]

P ( A ∩ B ∩ C) = 1/365
P(A) = 1/365, P(B)= 1/365, P(C) = 365
If events A,B and C are independed then P (A ∩ B ∩ C) = P (A) P(B) P(C) must be true,
From the probabilities we have
1/365≠ 1/365 * 1/365 * 1/365
Thus, events A,B, C are not independent.

3 0

1 year ago

Use Stokes' Theorem to evaluate S curl F · dS. F(x, y, z) = x2 sin(z)i + y2j + xyk, S is the part of the paraboloid z = 9 − x2 −

Korolek [52]

The vector field

$\vec F(x,y,z)=x^2\sin z\,\vec\imath+y^2\,\vec\jmath+xy\,\vec k$

has curl

$\nabla\times\vec F(x,y,z)=x\,\vec\imath+(x^2\cos z-y)\,\vec\jmath$

Parameterize $S$ by

$\vec s(u,v)=x(u,v)\,\vec\imath+y(u,v)\,\vec\jmath+z(u,v)\,\vec k$

where

$\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=(9-u^2)\end{cases}$

with $0\le u\le3$ and $0\le v\le2\pi$ .

Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k$

Then by Stokes' theorem we have

$\displaystyle\int_{\partial S}\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^3(\nabla\times\vec F)(\vec s(u,v))\cdot\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv$