Ask question

Ask question

All categories

3 years ago

7

Ms. Allen makes beaded purses. She used 4,725 beads to make 3 identical purses (they are exactly the same and use the same numbe

r of beads). How many beads per purse did Ms. Allen use?

1 answer:

solong [7]3 years ago

5 0

4,725 divided by 3 equals 1,575

You might be interested in

At lunch, Marc bought 2 quesadillas and 2 tacos for $10.16, and Jason paid $11.84 for 5 tacos and 1 quesadilla. What was the cos

mash [69]

Answer:

At lunch, Marc bought 2 quesadillas and 2 tacos for $10.16, and Jason paid $11.84 for 5 tacos and 1 quesadilla. What was the cost of each taco and quesadilla?

I want to know only this question because I am not also so so I tell me

4 0

3 years ago

How do you simplify 1/5 x 22

svetlana [45]

Answer:

The person above is correct

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

May I please receive help?

weeeeeb [17]

Answer:

you may

2x+15+4x+21=90 simplified is 6x+36=90

m<a=33º

m<b=90º

m<c=57º

Step-by-step explanation:

6x+36=90

6x=54

x=9

2(9)+15=33

4(9)+21=57

4 0

2 years ago

dalvyx [7]

Using the Poisson distribution, it is found that there is a 0.507 = 50.7% probability that the bird feeder will be visited by at most 5 birds in a 45 minute period during daylight hours.

<h3>What is the Poisson distribution?</h3>

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

The parameters are:

x is the number of successes

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Considering the average of 15 birds every 2 hours during daylight hours, the mean for a 45-minute period is given by:

$\mu = 15 \times \frac{45}{120} = 5.625$

The probability that the bird feeder will be visited by at most 5 birds in a 45 minute period during daylight hours is given by:

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

In which:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-5.625}(5.625)^{0}}{(0)!} = 0.004$

$P(X = 1) = \frac{e^{-5.625}(5.625)^{1}}{(1)!} = 0.02$

$P(X = 2) = \frac{e^{-5.625}(5.625)^{2}}{(2)!} = 0.057$

$P(X = 3) = \frac{e^{-5.625}(5.625)^{3}}{(3)!} = 0.107$

$P(X = 4) = \frac{e^{-5.625}(5.625)^{4}}{(4)!} = 0.150$

$P(X = 5) = \frac{e^{-5.625}(5.625)^{5}}{(5)!} = 0.169$

Then:

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.004 + 0.02 + 0.057 + 0.107 + 0.15 + 0.169 = 0.507$

0.507 = 50.7% probability that the bird feeder will be visited by at most 5 birds in a 45 minute period during daylight hours.

More can be learned about the Poisson distribution at brainly.com/question/13971530

#SPJ1

3 0

1 year ago

The measure of HMQ is (4x – 20)° ?

Kipish [7]

Answer:

A

Step-by-step explanation:

7 0

3 years ago