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professor190 [17]

2 years ago

12

HELP PLS MY MOM WILL BEAT ME HELP THXS

2 answers:

Tasya [4]2 years ago

8 0

Answer:

Solution is the missing word.

Step-by-step explanation:

Pepsi [2]2 years ago

7 0

Answer:

Solution

Step-by-step explanation:

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How hard is 6th grade math?

Ulleksa [173]

Answer:

super easy i can help you whenever

Step-by-step explanation:

im in high and i had As in 6th

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2 years ago

There are 294 boys and 322 girls in the hill school. how many students are in the school.

lys-0071 [83]

I think it's 616 that what I got but I'm not sure

7 0

3 years ago

lana66690 [7]

Answer:

see explanation

Step-by-step explanation:

Given the recursive formula

$a_{n}$ = $\frac{1}{2}$ $a_{n-1}$ with a₁ = 1, then

a₂ = $\frac{1}{2}$ a₁ = $\frac{1}{2}$ × 1 = $\frac{1}{2}$

a₃ = $\frac{1}{2}$ a₂ = $\frac{1}{2}$ × $\frac{1}{2}$ = $\frac{1}{4}$

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3 years ago

What is the period of the secant function

Alla [95]

Answer: C 2pi

Step-by-step explanation:

5 0

3 years ago

A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement

dsp73

Answer:

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

$P(Same) = \frac{67}{210}$

Step-by-step explanation:

Given (Omitted from the question)

$Red = 7$

$White = 9$

$Blue = 5$

Solving (a): $P(First\ White\ and\ Second\ Blue)$

This is calculated using:

$P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)$

$P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}$

<em>We used Total - 1 because it is a probability without replacement</em>

So, we have:

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}$

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}$

$P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}$

$P(First\ White\ and\ Second\ Blue) = \frac{45}{420}$

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

Solving (b) $P(Same)$

This is calculated as:

$P(Same) = P(First\ Blue\ and Second\ Blue)\$ $or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)$

$P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})$

$P(Same) = (\frac{5}{21} * \frac{4}{20})+(\frac{7}{21} * \frac{6}{20})+(\frac{9}{21} * \frac{8}{20})$

$P(Same) = \frac{20}{420}+\frac{42}{420} +\frac{72}{420}$

$P(Same) = \frac{20+42+72}{420}$

$P(Same) = \frac{134}{420}$

$P(Same) = \frac{67}{210}$

6 0

2 years ago