Ask question

Ask question

All categories

4 years ago

14

At a bake sale there were 90 items sold total. If 42 of the items sold were cookies and the rest were brownies what is the ratio

of brownies sold to cookies sold?

1 answer:

kari74 [83]4 years ago

5 0

Firstly, you do 90 - 42 to get the amount of brownies sold, which is 48. Then, you put 42 and 48 into a ratio like 42:48. After that, all you need to do is simplify to get 7:8.

You might be interested in

An article suggests that a poisson process can be used to represent the occurrence of structural loads over time. suppose the me

kirill115 [55]

Answer:

a) $\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

b) $P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

c) $e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution"

Solution to the problem

Let X our random variable who represent the "occurrence of structural loads over time"

For this case we have the value for the mean given $\mu = 0.5$ and we can solve for the parameter $\lambda$ like this:

$\frac{1}{\lambda} = 0.5$

$\lambda =2$

So then $X(t) \sim Poi (\lambda t)$

X follows a Poisson process

Part a

For this case since we are interested in the number of loads in a 2 year period the new rate would be given by:

$\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

Part b

For this case we want the following probability:

$P(X(2) >6)$

And we can use the complement rule like this

$P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

And we can solve this like this using the masss function:

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

Part c

For this case we know that the arrival time follows an exponential distribution and let T the random variable:

$T \sim Exp(\lambda=2)$

The probability of no arrival during a period of duration t is given by:

$f(T) = e^{-\lambda t}$

And we want to find a value of t who satisfy this:

$e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

3 0

4 years ago

I need help pleaseee):

ankoles [38]

Answer:

Step-by-step explanation:

7 0

3 years ago

Please help will mark brainly!

Marina86 [1]

Hey man I got u. The answer for this question is B. Peace

3 0

3 years ago

Read 2 more answers

Which equation can be used to represent six added to twice the sum of a numbee and four is equal to one half of the difference o

Ronch [10]

Answer: $6 + 2 (x+4)$ = $\frac{3}{2}$ $(3-x)$

Step-by-step explanation:

let the number be x , then :

six added to twice the sum of a number and four means :

$6 + 2 (x+4)$

one half of the difference of three and the number means :

$\frac{3}{2}$ $(3-x)$

combining the two , we have

$6 + 2 (x+4)$ = $\frac{3}{2}$ $(3-x)$

6 0

4 years ago

the ratio of cats and dogs at the city pound is 92 to 144.which statement descrbes this relationship?

Alexeev081 [22]

Answer:

There are 23 cats for every 36 dogs at the city pound

Step-by-step explanation:

This means that for every group of 92 cats at the city compound there is a group of 144 dogs. So it also means that for every group of 23 cats there is a group of 36 dogs.

92 to 144 or 92/144 = 23/26 or 23 to 36

<h2>Hope This Helps Out</h2>

5 0

2 years ago

Read 2 more answers