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Leokris [45]
4 years ago
10

Matthew jarred 10 liters of jam after 5 days. How many days does Matthew need to spend making jam if he wants to jar 18 liters o

f jam in all? Solve using unit rates
Mathematics
1 answer:
guajiro [1.7K]4 years ago
7 0

Answer:

  9 days

Step-by-step explanation:

Ordinarily, a rate that involves time will have the time unit in the denominator. Here, we have time as the dependent variable, so we want the time unit in the numerator for this problem. Then the "unit rate" is ...

  days/liter = (5 days)/(10 liters) = (1/2) day/liter

Then for 18 liters, we multiply by the unit rate to get days:

  (18 liters)(1/2 day/liter) = 9 days

It will take Matthew 9 days to jar 18 liters if he maintains his rate.

_____

The other way to write this unit rate is ...

  1/(1/2 day/liter) = 2 liters/day

Then to find days, we'd have to divide the desired volume by this rate:

  (18 liters)/(2 liters/day) = 9 days

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3 years ago
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Y=-3x+3
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3 years ago
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7 0
3 years ago
Exercise 5.2. Suppose that X has moment generating function
soldi70 [24.7K]

Answer:

a) Mean, E(X) = - 0.5

Variance = = 9.25

b) M_{X}(t)=E(e^{tX})

or

⇒ =\sum e^{tx}p(x)

⇒ =\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}

Step-by-step explanation:

Given:

moment generating function  of X as:

MX(t) = \frac{1}{2} + \frac{1}{3}e^{-4t} + \frac{1}{6} e^{5t}

a)  Now

Mean, E(X) = M_{X}'(t=0)

Thus,

M_{X}'(t)=\frac{1}{3}(-4)e^{-4t}+\frac{1}{6}(5)e^{5t}

or

M_{X}'(t)=\frac{-4}{3}e^{-4t}+\frac{5}{6}e^{5t}

also,

E(X^{2})=M_{X}''(t=0)

Thus,

M_{X}''(t)=\frac{-4}{3}(-4)e^{-4t}+\frac{5}{6}(5)e^{5t}

or

M_{X}''(t)=\frac{16}{3}e^{-4t}+\frac{25}{6}e^{5t}

Therefore,

Mean, E(X) = M_{X}'(t=0)=\frac{-4}{3}e^{-4(0)}+\frac{5}{6}e^{5(0)}

or

Mean, E(X) = - 0.5

and

E(X^{2})=M_{X}''(t=0)=\frac{16}{3}e^{-4(0)}+\frac{25}{6}e^{5(0)}

or

E(X^{2}) = 9.5

also,

Variance(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

b) Now,

Let f(x) be the PMF of X

Thus,

M_{X}(t)=E(e^{tX})

or

⇒ =\sum e^{tx}p(x)

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Therefore,

at x = 0, P(x) = \frac{1}{2}

at x= - 4 ,P(x) = \frac{1}{3}

at x = 5, P(x) = \frac{1}{6}

Thus,

E(X) =\sum xP(x)=0(\frac{1}{2})+(-4)(\frac{1}{3})+5(\frac{1}{6})

or

E(X) = - 0.5

also,E(X^{2})=\sum x^{2}P(x)=0^{2}(\frac{1}{2})+(-4)^{2}(\frac{1}{3})+5^{2}(\frac{1}{6})

E(X^{2})  = 9.5

Hence,

Var(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

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4 years ago
Help me please !!?????????? Question 4
Ne4ueva [31]

from question d = 16t^2 = 56-4t

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=》 (4t-7)(t+2) =0

=》 t = 1.75 seconds

4 0
3 years ago
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