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Fiesta28 [93]
2 years ago
15

Solve for u . 3u+11=41

Mathematics
2 answers:
Vinil7 [7]2 years ago
5 0
U is 10. If you subtract 11 first to get the variable and it's coefficient on its own, you see that 3u=30. To get the u on its own, you divide both sides by three. This gives you u=10
Hunter-Best [27]2 years ago
3 0
3u+11=41
-11 -11
3u=30
÷3 ÷3
u=10

I hope it helps!
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stich3 [128]

Answer:

C

Step-by-step explanation:

Because you should be subtract each other by 20x to transfer from side to another side.

4 0
3 years ago
Read 2 more answers
What is the value of x in the equation 6(x +11)= -300?
Kay [80]

Answer:

-61

Step-by-step explanation:

We use the distributive property to distribute 6 to both x and 11. This gives 6x+66=-300. Subtracting both sides by 66 gives 6x=-366. Dividing both sides gives x=-61.

8 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
How do you add 323+412 in base 5
Alenkasestr [34]

Answer:

Addition of 323+412 = 1240 at base 5

Step-by-step explanation:

We have to add 323+412 on base 5

As we have to use base 5 for addition so we can use only numbers 0,1,2,3.4 and 5 as we use 0,1,2,3,4,5,6,7,8,9,10 in decimal when base is 10

Means in addition of sum is greater than 5 then we have to take carry as in base 10 we take carry if the sum is greater than 10

So sum 323+412 = 1240

8 0
3 years ago
The picture frames shown are both squares. The area of the smaller frame is 1/4 the area of the larger frame. Find the side leng
baherus [9]
Area of smaller frame =1/4×36=9sq.in
side of smaller frame =
\sqrt{9  }  = 3in
side of larger frame
=  \sqrt{36}  = 6in
4 0
3 years ago
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