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3 years ago

X3 - 10 = And x is 6

Mathematics

2 answers:

Marysya12 [62]3 years ago

7 0

X = 18 think of it this way what can you multiply by 3 that if you take 10 away would equal the answer to your equation

shtirl [24]3 years ago

4 0

Answer:8

Step-by-stop explanation: (6) x 3 will be 18 - 10 = 8

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Circle O has a circumference of approximately 250 ft

Varvara68 [4.7K]

Answer:

80 feet

Step-by-step explanation:

c=d $\pi$

Substitute 250 in for circumference

250= $\pi$ d

Divide by $\pi$ or 3.14

d=79.617...

So, the diameter is approximately 80 feet

Hope this helps! :)

8 0

3 years ago

Find DE. F) 3 G) 6 H) 9 J) 12

mestny [16]

Answer:

Step-by-step explanation:

F because im guessing and im taking a tes

6 0

2 years ago

In the expression 3x-10 which would be the constant

ArbitrLikvidat [17]

Answer:

-10

Step-by-step explanation:

The constant is the part of the algebraic function or expression that does not change. In this case, as x changes, the constant -10 does not. Thus, -10 is the constant.

4 0

2 years ago

Will give brainliest!

fiasKO [112]

Let's give some distances
100 up 100 flat 100 down meters
then return 100 up 100 flat 100 down

time required (distance/rate = time ) 200/100 + 200/120 + 200 / 150 = 5 hr

distance 600

distance / time = average speed = 600/5 = 120 m/s

5 0

2 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