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3 years ago

A recipe for a dessert for 6 people requires 120g of sugar. How much sugar would I need for a dessert 9 people?

2 answers:

4 0

(120/6) = 20 grams of sugar per person.
(9 people times grams of sugar per person.
180 grams of sugar for a recipe serving 9 People.

zzz [600]3 years ago

3 0

Answer:

180g of sugar

Step-by-step explanation:

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81/450 as a percent equals 18%

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3 years ago

If f(x)=x+6 and g(x)=x^4 find G(F(x))

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2 years ago

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(39 points PLZ HELP)Find the sum of a finite geometric series. A ball is dropped from a height of 10 meters. Each time it bounce

erastovalidia [21]

So we need to find the sum of the first 5 terms.

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Plugging in the terms that we know...

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3 years ago

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Find the general solution of y\"+9y=t^2*e^3t+6.

Fofino [41]

We are given with the equation y"+ 9 y = t^2 * e^3 t + 6 and asked to determine the general solution to this equation. we convert the equation into r^2 + 9 = 0. The solution to this equation is r = +/- 3i. This solution converted to trigonometric function is equivalent to y= C1 cos3t + C2 sin3t.

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3 years ago

The number of people arriving for treatment at an emergency room can be modeled by aPoisson process with a rate parameter of 5/h

saveliy_v [14]

Answer:

(a) The probability of having exactly four arrivals during a particular hour is 0.1754.

(b) The probability that at least 3 people arriving during a particular hour is 0.7350.

Step-by-step explanation:

(a) If the arrivals can be modeled by a Poisson process, with λ = 5/hr, the probability of having exactly four arrivals during a particular hour is:

$P(X=4)=\frac{\lambda^{X}*e^{-\lambda} }{X!} =\frac{5^{4}*e^{-5} }{4!}=\frac{625*0.006737947}{24} =\frac{4.211}{24}=0.1754$

The probability of having exactly four arrivals during a particular hour is 0.1754.

(b) The probability that at least 3 people arriving during a particular hour can be written as

$P(X>3)=1-P(X\leq3)=1- (P(0)+P(1)+P(2)+P(3))$

Using

$P(X)=\frac{\lambda^{X}*e^{-\lambda} }{X!}$

We get

$P(X>3)=1-P(X\leq3)=1- (P(0)+P(1)+P(2)+P(3))\\P(X>3)=1-(0.0067+ 0.0337+ 0.0842 + 0.1404 )\\P(X>3)=1-0.2650=0.7350\\$

The probability that at least 3 people arriving during a particular hour is 0.7350.

$EV=\lambda*t=5*0.75=3.75$

8 0

2 years ago