All categories

3 years ago

HELP PLEASE

1 answer:

5 0

1 worker=1/5 grass per hour
6 workers

number of workers times work each worker does times time needed=total work done
5 times ? times 1=1 lawn
?=1/5 lawn
1 worker does 1/5 lawn in 1hour

6 workers
work is 1/5
time=?
total work=1

6 times 1/5 times ? =1
6/5 times ?=1
times 5/6 both sides
?=5/6

it takes 5/6 hour or 50 minutes

You might be interested in

The table shows the results of spinning a four colored spinner 50 times. Find the experimental probability and express it as a d

gogolik [260]

Red is 20 and you spin it 50 times so 2/5 chance is red so that makes 3/5 chance not red

5 0

3 years ago

QUICK HELP!!! How manny differences can you spot? 8 6 7 4 3 5

Eduardwww [97]

Answer:

i can only see 3

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

The scale on a map shows that 200 miles is represented by 5 inches on a map which proportion can be used to find the distance of

IRISSAK [1]

Answer:

gzhausiaoa

Step-by-step explanation:

hshshsshhehehehehhdhdhdhdhhdhdhdfhfh

3 0

3 years ago

Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie

ludmilkaskok [199]

Answer:

$\lambda \geq 6.63835$

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that $X \sim Poisson(\lambda)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda$

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

$P(X\geq 2)=1-P(X$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}$

$P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}$

And replacing we have this:

$P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]$

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)$

And we want this probability that at least of 99%, so we can set upt the following inequality:

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99$

And now we can solve for $\lambda$

$0.01 \geq e^{-\lambda}(1+\lambda)$

Applying natural log on both sides we have:

$ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)$

$ln(0.01) \geq -\lambda+ln(1+\lambda)$

$\lambda-ln(1+\lambda)+ln(0.01) \geq 0$

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

$x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}$