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3 years ago

Pls help !!!!!!!!!!!!!!!!!!!lllllllllllllllllllllllll!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

adoni [48]3 years ago

3 0

Can you take the picture closer?

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How many models of 100 do you need to model 3200 explain

rusak2 [61]

32 models need to make model of 3200.

Given that a 1 model contain 100.

Two series of numbers, usually empirical data, that are proportional or proportional if their respective elements are in constant proportion, called the scaling factor or the rate constant.

One model has 100 elements.

Now, we have to find how many model contains 3200 elements.

So, 1 model=100 elements

n model =3200 elements

We will write this in proportion as

1/n=100/3200

Applying the cross multiply, we get

3200×1=n×100

Divide both sides with 100, we get

3200/100=100n/100

3200/100=n

32=n

Hence, the 32 models contain 3200 elements when one contain 100 elements.

Learn more about proportional from here brainly.com/question/23536327

#SPJ9

4 0

1 year ago

Michael has two more apples than Ann does. Together, Michael and Ann have no more than 9 apples. If Ann has at least one apple,

fiasKO [112]

Answer:

ann has 1 apple and Michael has 8 apples

6 0

3 years ago

How do you graph a linear function to a line graph

anygoal [31]

Answer:

use point slope form

y=mx+b

8 0

3 years ago

I really need help on this!

Anvisha [2.4K]

103-12=91 so 91 i think

6 0

3 years ago

Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2

leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where $\lambda = \frac{10.2}{200} = 0.051$

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

$\frac{1}{4} \times 200 = 50$

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

$P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004$

c)The expected number of grams to be eaten before encountering the first fragments :

$E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 gram$ s

7 0

3 years ago