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Sauron [17]
2 years ago
12

Please help again ASAP!!! I appreciate answers!

Mathematics
1 answer:
Natalija [7]2 years ago
5 0

Answer

The answer would be the second one!

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Surface area of solid cuboid 8cm 6cm 2cm
strojnjashka [21]

Answer:

152 \text{cm}^2

Step-by-step explanation:

We know that the total surface area of a cuboid is the sum of the  surface area of all six of its faces.

We also know that each opposing side has the same surface area. Thus the total surface area is equal to:

2*(8*6+8*2+6*2)\\\\=2*(48+16+12)\\=2*76\\=152(\text{cm}^2)

6 0
2 years ago
The equation of a line is 8x+2y=10. Calculate the slope of the line.
emmainna [20.7K]

Answer:

that is the solution to the question

6 0
3 years ago
Which of these is a step in constructing an inscribed equilateral triangle using technology?
lyudmila [28]
I think that the answer is B. I tried to read more about the question to understand it more but got nothing out of it. 
5 0
3 years ago
Draw a quadrilateral that does not belong then explain why
tatyana61 [14]

Answer:

a kite

Step-by-step explanation:

6 0
3 years ago
Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand
dybincka [34]

Answer: Part a) P(a)=\frac{1}{\binom{r+w}{r}}

part b)P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = P(E)=\frac{Favourablecases}{TotalCases}

For part a)

Probability that a red ball is drawn in first attempt = P(E_{1})=\frac{r}{r+w}

Probability that a red ball is drawn in second attempt=P(E_{2})=\frac{r-1}{r+w-1}

Probability that a red ball is drawn in third attempt = P(E_{3})=\frac{r-2}{r+w-1}

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = P(E_{i})=\frac{r-i}{r+w-i}

Thus the probability that events E_{1},E_{2}....E_{i} occur in succession is

P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...

Thus P(E)=\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!

Thus our probability becomes

P(E)=\frac{1}{\binom{r+w}{r}}

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at (r+1)^{th} draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals \binom{r}{r-1}

Thus the probability becomes P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}

Thus required probability of case b becomes P(E)+ P(E_{i})

= P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\

7 0
3 years ago
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