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3 years ago

Someone Help Me Please ? , It’s Probability

Mathematics

1 answer:

olga nikolaevna [1]3 years ago

3 0

1. correct
2. same as 1
3. 2/5, 2:5, .40, 40%
4. 3/5, 3:5, 0.60, 60%

Hope this helped☺☺

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What is the 12th term of the sequence -2,-4,-6,...........-100?

Margaret [11]

Answer:

-24

Step-by-step explanation:

You are just counting by two's.

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2 years ago

Read 2 more answers

What is the product 4n/4n-4 n-1/n+1

olga_2 [115]

4n/4n-4 × n-1/n+1
= 4n(n-1)/(4n-4)(n+1)
= (4n^2 - 4n)/(4n^2 + 4n - 4n-4)
= (4n^2 - 4n)/(4n^2 - 4)
= 4(n^2 - n)/4(n^2 - 1)
=(n^2 - n)/(n^2 - 1)

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3 years ago

Please help.<br> If you could put your work in the answer that would<br> Be greatly appreciated.

AleksandrR [38]

Answer:

38°

because it is equal.

GJ=21

JH=21

GJ=JH

∠GFJ =∠HFJ

∠GFJ =38°

∠HFJ=38°

6 0

3 years ago

Pls help!!!!The probability of hitting a target on a single trial is 40%,Find the probability of hitting the same target 5/8 tim

azamat

Probability hit=40%=4/10
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probability= p(hit)+p(hit)+p(hit)+p(hit)+p(hit)+p (no hit)+p (no hit)+p (no hit)
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3 years ago

Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N defini

Nat2105 [25]

By inspection, it's clear that the sequence must converge to $\dfrac32$ because

$\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32$

when $n$ is arbitrarily large.

Now, for the limit as $n\to\infty$ to be equal to $\dfrac32$ is to say that for any $\varepsilon>0$ , there exists some $N$ such that whenever $n>N$ , it follows that

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|$

From this inequality, we get

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}$
$\implies|2n-1|>\dfrac5{2\varepsilon}$
$\implies2n-1\dfrac5{2\varepsilon}$
$\implies n\dfrac12+\dfrac5{4\varepsilon}$

As we're considering $n\to\infty$ , we can omit the first inequality.

We can then see that choosing $N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil$ will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that $N\in\mathbb N$ .

6 0

3 years ago