All categories

3 years ago

PLZ HELP FAST

Mathematics

2 answers:

natka813 [3]3 years ago

7 0

Answer:

11.81

Step-by-step explanation:

5 1/4 per 20 min (1/3 of an hour)

5.25 (5 1/4) * 3 = 15.75 pounds per hour

$0.75 earned per pound

So, 15.75 * $0.75 = $11.81

Harrizon [31]3 years ago

7 0

Answer:

Berries per ⅓ hour = 5¼ pounds =21/4

Per pound = $0.75

Berries to be picked in one hour = 21/4 × 3 = 15¾ pounds

Total money made in 1 hr= 15¾ × 0.75 = $11.81

You might be interested in

Suppose each of 12 players rolls a pair of dice 3 times. Find the probability that at least 4 of the players will roll doubles a

polet [3.4K]

Answer:

Our answer is 0.8172

Step-by-step explanation:

P(doubles on a single roll of pair of dice) =(6/36) =1/6

therefore P(in 3 rolls of pair of dice at least one doubles)=1-P(none of roll shows a double)

=1-(1-1/6)3 =91/216

for 12 players this follows binomial distribution with parameter n=12 and p=91/216

probability that at least 4 of the players will get “doubles” at least once =P(X>=4)

=1-(P(X<=3)

=1-((₁₂ C0)×(91/216)⁰(125/216)¹²+(₁₂ C1)×(91/216)¹(125/216)¹¹+(₁₂ C2)×(91/216)²(125/216)¹⁰+(₁₂ C3)×(91/216)³(125/216)⁹)

=1-0.1828

=0.8172

7 0

3 years ago

Our faucet is broken, and a plumber has been called. The arrival time of the plumber is uniformly distributed between 1pm and 7p

Ymorist [56]

Answer:

$E(A+B) = E(A)+E(B)=4+0.5 =4.5 hours$

$Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2$

Step-by-step explanation:

Let A the random variable that represent "The arrival time of the plumber ". And we know that the distribution of A is given by:

$A\sim Uniform(1 ,7)$

And let B the random variable that represent "The time required to fix the broken faucet". And we know the distribution of B, given by:

$B\sim Exp(\lambda=\frac{1}{30 min})$

Supposing that the two times are independent, find the expected value and the variance of the time at which the plumber completes the project.

So we are interested on the expected value of A+B, like this

$E(A +B)$

Since the two random variables are assumed independent, then we have this

$E(A+B) = E(A)+E(B)$

So we can find the individual expected values for each distribution and then we can add it.

For ths uniform distribution the expected value is given by $E(X) =\frac{a+b}{2}$ where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:

$E(A)=\frac{1+7}{2}=4 hours$

The expected value for the exponential distirbution is given by :

$E(X)= \int_{0}^\infty x \lambda e^{-\lambda x} dx$

If we use the substitution $y=\lambda x$ we have this:

$E(X)=\frac{1}{\lambda} \int_{0}^\infty y e^{-\lambda y} dy =\frac{1}{\lambda}$

Where X represent the random variable and $\lambda$ the parameter. If we apply this formula to our case we got:

$E(B) =\frac{1}{\lambda}=\frac{1}{\frac{1}{30}}=30min$

We can convert this into hours and we got E(B) =0.5 hours, and then we can find:

$E(A+B) = E(A)+E(B)=4+0.5 =4.5 hours$

And in order to find the variance for the random variable A+B we can find the individual variances:

$Var(A)= \frac{(b-a)^2}{12}=\frac{(7-1)^2}{12}=3 hours^2$

$Var(B) =\frac{1}{\lambda^2}=\frac{1}{(\frac{1}{30})^2}=900 min^2 x\frac{1hr^2}{3600 min^2}=0.25 hours^2$

We have the following property:

$Var(X+Y)= Var(X)+Var(Y) +2 Cov(X,Y)$

Since we have independnet variable the Cov(A,B)=0, so then:

$Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2$

3 0

3 years ago

Is 2/3 greater than 10/18

SVEN [57.7K]

2/3 is greater than 10/18

4 0

2 years ago

Read 2 more answers

Show how you can use repeated subtraction to find 84 divided by 6

Luba_88 [7]

84-14=70, 70-14=56, 56-14=42, 42-14=28, 28-14=14, 14-14=0. Hope this helped!!

8 0

3 years ago

Read 2 more answers

A schools running track is 200m long (in an oval shape). How many times must a runner go around the track to run 1.6km?

LUCKY_DIMON [66]

It is 8 laps (I think)

8 0

2 years ago

Read 2 more answers