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3 years ago

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Jacqueline has a bag containing six black checkers and four red checkers. What is the probability of her selecting two black che

ckers if she does replace the first checker she pulls out of the bag before selecting the second checker?

1 answer:

Marat540 [252]3 years ago

7 0

I think the probability is 1 because they’re mixed

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1. What is a real world example when you would have to find the circumference of something? Explain.

nikitadnepr [17]

A ferris wheel
A construction worker whose making a ferris wheel would need to know and measure the circumference.
A theme park manger would need to find the circumference to know if the ferris wheel is too big to go into the park.
Hope that helps!!

4 0

3 years ago

Calculus hw, need help asap with steps.

nikdorinn [45]

Answers are in bold

S1 = 1

S2 = 0.5

S3 = 0.6667

S4 = 0.625

S5 = 0.6333

=========================================================

Explanation:

Let $f(n) = \frac{(-1)^{n+1}}{n!}$

The summation given to us represents the following

$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}=\sum_{n=1}^{\infty} f(n)\\\\\\\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}=f(1) + f(2)+f(3)+\ldots\\\\$

There are infinitely many terms to be added.

-------------------

The partial sums only care about adding a finite amount of terms.

The partial sum $S_1$ is the sum of the first term and nothing else. Technically it's not really a sum because it doesn't have any other thing to add to. So we simply say $S_1 = f(1) = 1$

I'm skipping the steps to compute f(1) since you already have done so.

-------------------

The second partial sum is when things get a bit more interesting.

We add the first two terms.

$S_2 = f(1)+f(2)\\\\S_2 = 1+(-\frac{1}{2})\\\\S_2 = \frac{1}{2}\\\\S_2 = 0.5\\\\\\$

The scratch work for computing f(2) is shown in the diagram below.

-------------------

We do the same type of steps for the third partial sum.

$S_3 = f(1)+f(2)+f(3)\\\\S_3 = 1+(-\frac{1}{2})+\frac{1}{6}\\\\S_3 = \frac{2}{3}\\\\S_3 \approx 0.6667\\\\\\$

The scratch work for computing f(3) is shown in the diagram below.

-------------------

Now add the first four terms to get the fourth partial sum.

$S_4 = f(1)+f(2)+f(3)+f(4)\\\\S_4 = 1+(-\frac{1}{2})+\frac{1}{6}-\frac{1}{24}\\\\S_4 = \frac{5}{8}\\\\S_4 \approx 0.625\\\\\\$

As before, the scratch work for f(4) is shown below.

I'm sure you can notice by now, but the partial sums are recursive. Each new partial sum builds upon what is already added up so far.

This means something like $S_3 = S_2 + f(3)$ and $S_4 = S_3 + f(4)$

In general, $S_{n+1} = S_{n} + f(n+1)$ so you don't have to add up all the first n terms. Simply add the last term to the previous partial sum.

-------------------

Let's use that recursive trick to find $S_5$

$S_5 = [f(1)+f(2)+f(3)+f(4)]+f(5)\\\\S_5 = S_4 + f(5)\\\\S_5 = \frac{5}{8} + \frac{1}{120}\\\\S_5 = \frac{19}{30}\\\\S_5 \approx 0.6333$

The scratch work for f(5) is shown below.

7 0

2 years ago

What is the range of g(x)?

Jlenok [28]

Answer:

The function g(x) is defined for all real numbers x. The maximum value of the range is 4. The maximum value of the domain is 3. The range of g(x) is {y| –1 < y ≤ 4}.

Step-by-step explanation:

i don't know what your question is exactly but hope this helps.

8 0

3 years ago

What is the chance of rolling 4 a million time?

lys-0071 [83]

$\frac{1}{1000 \: 000}$

6 0

3 years ago

Simplify <img src="https://tex.z-dn.net/?f=%5Csqrt%20-25" id="TexFormula1" title="\sqrt -25" alt="\sqrt -25" align="absmiddle" c

garik1379 [7]

Answer:

5<em>i</em>

Step-by-step explanation:

= $\sqrt{-1 * 25}$
= $\sqrt{-1}\sqrt{25}$
= $\sqrt{-1}$ * 5

Rule: $\sqrt{-1}$ = <em>i</em>

So you get <em>i </em>* 5<em> </em>=<em> </em>5<em>i</em>

5 0

3 years ago