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2 years ago

7

Jasmine earns $36 for 4 hours of baby-sitting. She charges a constant hourly rate. Which table correctly shows the amount Jasmin

e earns baby-sitting for different numbers of hours?

2 answers:

Vadim26 [7]2 years ago

5 0

I am pretty sure it is a because 2 is half of 4 and 18 is half of 36.

kifflom [539]2 years ago

3 0

Answer:

A

Step-by-step explanation:

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Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 1 + sec

masha68 [24]

Using the washer method, the volume is given by the integral

$\displaystyle\pi\int_{-\pi/3}^{\pi/3}\bigg((3-1)^2-((1+\sec x)-1)^2\bigg)\,\mathrm dx=2\pi\int_0^{\pi/3}(4-\sec^2x)\,\mathrm dx$

where 3 - 1 = 2 is the distance from <em>y</em> = 3 to the axis of revolution, and similarly (1 + sec(<em>x</em>)) - 1 = sec(<em>x</em>) is the distance from <em>y</em> = 1 + sec(<em>x</em>) to the axis. The integrand is symmetric about <em>x</em> = 0, so the integral "folds" in on itself, and the integral from -π/3 to π/3 is twice the integral from 0 to π/3.

So the volume is

$\displaystyle2\pi\int_0^{\pi/3}(4-\sec^2x)\,\mathrm dx=2\pi(4x-\tan x)\bigg|_0^{\pi/3}=\boxed{\dfrac{8\pi^2}3-2\pi\sqrt3}$

4 0

2 years ago

let x = the amoun of raw sugar in tons a procesing plant is a sugar refinery process in one day . suppose x can be model as expo

anygoal [31]

Answer:

The answer is below

Step-by-step explanation:

A sugar refinery has three processing plants, all receiving raw sugar in bulk. The amount of raw sugar (in tons) that one plant can process in one day can be modelled using an exponential distribution with mean of 4 tons for each of three plants. If each plant operates independently,a.Find the probability that any given plant processes more than 5 tons of raw sugar on a given day.b.Find the probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day.c.How much raw sugar should be stocked for the plant each day so that the chance of running out of the raw sugar is only 0.05?

Answer: The mean (μ) of the plants is 4 tons. The probability density function of an exponential distribution is given by:

$f(x)=\lambda e^{-\lambda x}\\But\ \lambda= 1/\mu=1/4 = 0.25\\Therefore:\\f(x)=0.25e^{-0.25x}\\$

a) P(x > 5) = $\int\limits^\infty_5 {f(x)} \, dx =\int\limits^\infty_5 {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_5=e^{-1.25}=0.2865$

b) Probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day can be solved when considered as a binomial.

That is P(2 of the three plant use more than five tons) = C(3,2) × [P(x > 5)]² × (1-P(x > 5)) = 3(0.2865²)(1-0.2865) = 0.1757

c) Let b be the amount of raw sugar should be stocked for the plant each day.

P(x > a) = $\int\limits^\infty_a {f(x)} \, dx =\int\limits^\infty_a {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_a=e^{-0.25a}$

But P(x > a) = 0.05

Therefore:

$e^{-0.25a}=0.05\\ln[e^{-0.25a}]=ln(0.05)\\-0.25a=-2.9957\\a=11.98$

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2 years ago

"You are going to make a trash bin with a square base , vertical sides and no top. If you have 108 ft2 of material available, de

kotykmax [81]

<h2>Dimension of trash bin is 6 ft x 6 ft x 3ft .</h2>

Step-by-step explanation:

Let h be the height of box and a be the side of base square.

We have 108 ft² material.

Surface area of trash bin = 108 ft²

Surface area of trash bin = a² + 4 ah

a² + 4 ah = 108

4ah = 108 - a²

$h=\frac{108-a^2}{4a}$

Volume of trash bin, V = a²h

$V=a^2\times \frac{108-a^2}{4a}=27a-0.25a^3$

At maximum volume differentiation of volume will be zero

$dV=27da-0.25\times 3a^2da\\\\0=27da-0.75a^2da\\\\a^2=36\\\\a=6ft$

Surface area of trash bin = a² + 4 ah = 108

36 + 4 x 6 x h = 108

h = 3 ft

Dimension of trash bin is 6 ft x 6 ft x 3ft .

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Answer:

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Step-by-step explanation:

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