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4 years ago

What is 1000÷3872911

Mathematics

1 answer:

nasty-shy [4]4 years ago

7 0

Answer:

The answer is 0.00025820371!!!XD

Hope this helped!!!XD

Also, can i get brainliest please?

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What is the interquartile range

elena-14-01-66 [18.8K]

The answer is six because you have to subtract the upper and lower quartile which are 18 and 16 in this case

3 0

3 years ago

If f(x)= x/2 -2 and g(x) = 2x2 +x-3, find (f +g)(x).

Molodets [167]

Answer:

Step-by-step explanation:

(f+g)(x) means add f(x) and g(x) together:

(x/2)-2 + 2x^2 +x -3

collect all the like terms:

2x^2

(x/2) + x = 3/2x (explanation below)

-2 -3 = -5

x/2 + x can be turned to 1x/2 + 2/2x (2/2 = 1, so when you simplify it to 1x, it is still x)

6 0

3 years ago

5. Write 2/3 and 3/4 as equivalent fractions using a common denominator.

hichkok12 [17]

Answer:

8/12, 9/12

Step-by-step explanation:

2/3 = 8/12

3/4 = 9/12

4 0

3 years ago

Read 2 more answers

Plsss anyone help meee plsss

Volgvan

Answer:

yes what do you want help with

8 0

3 years ago

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

bulgar [2K]

Answer:

a) 20.61%

b) 21.82%

c) 42.36%

d) 4 withdrawals

Step-by-step explanation:

This situation can be modeled with a binomial distribution, where p = probability of “success” (completing the course) equals 80% = 0.8 and the probability of “failure” (withdrawing) equals 0.2.

So, the probability of exactly k withdrawals in 20 cases is given by

$\large P(20;k)=\binom{20}{k}(0.2)^k(0.8)^{20-k}$

We are looking for

P(0;20)+P(0;1)+P(0;2) =

$\large \binom{20}{0}(0.2)^0(0.8)^{20}+\binom{20}{1}(0.2)^1(0.8)^{19}+\binom{20}{2}(0.2)^2(0.8)^{18}=$

0.0115292150460685 + 0.0576460752303424 + 0.136909428672063 = 0.206084718948474≅ 0.2061 or 20.61%

Here we want P(20;4)

$\large P(20;4)=\binom{20}{4}(0.2)^4(0.8)^{16}=0.218199402\approx 0.2182=21.82\%$

Here we need

$\large \sum_{k=4}^{20}P(20;k)=1-\sum_{k=1}^{3}P(20;k)$

But we already have P(0;20)+P(0;1)+P(0;2) =0.2061 and

$\large \sum_{k=1}^{3}P(20;k)=0.2061+P(20;3)=0.2061+0.205364 \approx 0.4236=42.36\%$

For a binomial distribution the <em>expectance </em>of “succeses” in n trials is np where p is the probability of “succes”, and the expectance of “failures” is nq, so the expectance for withdrawals in 20 students is 20*0.2 = <em>4 withdrawals.</em>

3 0

3 years ago