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viktelen [127]
3 years ago
15

Write the number in witch the digit 2 is one tenth the value of the digit 2 in 8.524.

Mathematics
1 answer:
spin [16.1K]3 years ago
4 0
Well the 2 in 8.524 would be 0.02 or 2 hundredths. Here's my explanation, the 8 is the whole number because it's before the decimal point, the 5 is the tenth or 0.5 and as you go further the number gets smaller so after the 5 is the 2 and that would be 2 hundredths or 0.02. So basically it's like going from tens, to hundreds, to thousands but only going in reverse order.
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Can someone help me with this surface area plz plz thanks
Lunna [17]
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4 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
Which list of numbers is in order from least to greatest?<br> 3/6 3/10 3/7
lapo4ka [179]
The list of numbers from least to greatest is 

3/10 , 3/7, 3/6

This is because the numerator stays the same across all three of these fractions.  However, the denominator is changing, and this proves to be a large difference.  

Think about it this way: If you cut a pie into ten pieces, then those slices are going to be smaller than if you cut the pie into seven or six pieces.

5 0
3 years ago
Read 2 more answers
The congruence theorem that can be used to prove ALON
Arisa [49]

Answer:

SSS is the congruence theorem that can be used to prove Δ LON  is congruent to Δ LMN ⇒ 1st answer

Step-by-step explanation:

Let us revise the cases of congruence

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  • ASA ⇒ 2 angles and the side whose joining them in the 1st Δ ≅ 2 angles and the side whose joining them in the 2nd Δ
  • AAS ⇒ 2 angles and one side in the 1st Δ ≅ 2 angles and one side in the 2nd Δ
  • HL ⇒ hypotenuse leg of the 1st right Δ ≅ hypotenuse leg of the 2nd right Δ  

In triangles LON and LMN

∵ LO ≅ LM ⇒ given

∵ NO ≅ NM ⇒ given

∵ LN is a common side in the two triangles

- That means the 3 sides of Δ LON are congruent to the 3 sides

   of Δ LMN

∴ Δ LON ≅ LMN ⇒ by using SSS theorem of congruence

SSS is the congruence theorem that can be used to prove Δ LON  is congruent to Δ LMN

3 0
3 years ago
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LenaWriter [7]

Answer:

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