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3 years ago

Can somebody plz help answer these probability questions correctly (only if u know how to do it ofc lol) thanks! :3

Mathematics

1 answer:

Serjik [45]3 years ago

4 0

Answer:

1/8

1/2

3/8

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How many significant figures are there In the number 10.76

VikaD [51]

Answer:

Number of Significant Figures: 4

The Significant Figures are 1 0 7 6

3 0

3 years ago

One grain of this sand approximately weighs 6 x 10^-6g.

emmasim [6.3K]

Answer:

1.2×10¹² grains

Step-by-step explanation:

1 grain = 6×10⁻⁶ g

7200 kg =
7200000 g =
7.2×10⁶ g =
7.2×10⁶ ÷ 6×10⁻⁶ =
(7.2÷6)×10⁶⁻ ⁽⁻⁶⁾ =
1.2×10¹² grains

4 0

3 years ago

The school yard has an area of 3564 ft.² the section that is in the shape of a triangle is used for picnic tables what is the ar

kicyunya [14]

x × 44ft =3564

× = 81

area of the section for a picnic

= 1/2 × 44 × 81

= 1782 ft^2

8 0

4 years ago

PLEASE HELPPPPPP EASY

lianna [129]

The answer to your question is D

5 0

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