Given O that is the center of the circle below, compare the quantity in column A with the quantity in column B.

What is the sum? StartFraction 3 y Over y squared 7 y 10 EndFraction StartFraction 2 Over y 2 EndFraction.

uranmaximum [27]

You can take LCM and then can add to see what's the sum.

The sum result of given expressions is given by: $\dfrac{5}{y+5}$

<h2>Given that:</h2>

To find sum of $\dfrac{3y}{y^2 + 7y + 10}$ and $\dfrac{2}{y+2}$

<h2>Finding LCM and summing:</h2>

You can take LCM and then can add to see whats the sum.

$\begin{aligned}\dfrac{3y}{y^2 + 7y + 10} + \dfrac{2}{y+2} &= \dfrac{3y}{(y+2)(y+5)} + \dfrac{2}{y+2}\\&= \dfrac{3y + 2(y+5)}{(y+2)(y+5)}\\&= \dfrac{5y+10}{(y+2)(y+5)}\\&= \dfrac{5(y+2)}{(y+2)(y+5)}\\&= \dfrac{5}{y+5}\\\end{aligned}$

Thus, the sum of given expressions is given by:

$\dfrac{5}{y+5}$

Learn more about algebraic sums here:

brainly.com/question/14964662

8 0

2 years ago

What expression can be used to add 1/2 plus 5/6

Vlad [161]

You can add the two fractions by turning 1/2 and 5/6 to have the same denominator.

Turn 1/2 into a fraction with a denominator of 6.
It would equal 3/6

5/6+ 3/6= 8/6

That simplifies to 1 1/3.

Hope this helps you!

Brainliest answer is always appreciated!

5 0

3 years ago

1960 $4,995

ruslelena [56]

Answer:

Average mean salary $=$26327.83$

Step-by-step explanation:

We have been given year and respective salaries in each year.

1960 $4,995

1970 $8,626

1980 $15,970

1990 $31,367

2000 $41,807

2010 $55,202

Now we nee do to determine the average mean salary for the six decades, 1960 – 2010.

So we just need to add all those six salaries and divide that sum by 6 to find the average mean salary.

Average mean salary $=\frac{(4995+8626+15970+31367+41807+55202)}{6}$

Average mean salary $=\frac{(157967)}{6}$

Average mean salary $=26327.83333333$

Hence final answer is approx:

Average mean salary $=$26327.83$

7 0

4 years ago

Read 2 more answers

Write the equation of the line given the points (2,0) and (3.-1)

Juli2301 [7.4K]

Answer:

y = -1x + 2

Step-by-step explanation:

We need to find y = mx + b

Use the following equation

$\frac{y_2-y_1}{x_2-x_1}$

Plug in the values

$\frac{-1-0}{3-2}$ = $\frac{-1}{1} = -1$

We have found the slope(m) which is -1

We have to find y-int b

Plug in any given x,y into the slope formula

-1 = -1(3) + b

-1 = -3 + b

2 = b

At last the formula is

y = -1x + 2

7 0

3 years ago

The computers of nine engineers at a certain company are to be replaced. Four of the engineers have selected laptops and the oth

Gala2k [10]

Answer:

(a) There are 70 different ways set up 4 computers out of 8.

(b) The probability that exactly three of the selected computers are desktops is 0.305.

(c) The probability that at least three of the selected computers are desktops is 0.401.

Step-by-step explanation:

Of the 9 new computers 4 are laptops and 5 are desktop.

Let X = a laptop is selected and Y = a desktop is selected.

The probability of selecting a laptop is = $P(Laptop) = p_{X} = \frac{4}{9}$

The probability of selecting a desktop is = $P(Desktop) = p_{Y} = \frac{5}{9}$

Then both X and Y follows Binomial distribution.

$X\sim Bin(9, \frac{4}{9})\\ Y\sim Bin(9, \frac{5}{9})$

The probability function of a binomial distribution is:

$P(U=k)={n\choose k}\times(p)^{k}\times (1-p)^{n-k}$

(a)

Combination is used to determine the number of ways to select <em>k</em> objects from <em>n</em> distinct objects without replacement.

It is denotes as: ${n\choose k}=\frac{n!}{k!(n-k)!}$

In this case 4 computers are to selected of 8 to be set up. Since there cannot be replacement, i.e. we cannot set up one computer twice or thrice, use combinations to determine the number of ways to set up 4 computers of 8.

The number of ways to set up 4 computers of 8 is:

${8\choose 4}=\frac{8!}{4!(8-4)!}\\=\frac{8!}{4!\times 4!} \\=70$

Thus, there are 70 different ways set up 4 computers out of 8.

(b)

It is provided that 4 computers are randomly selected.

Compute the probability that exactly 3 of the 4 computers selected are desktops as follows:

$P(Y=3)={4\choose 3}\times(\frac{5}{9})^{3}\times (1-\frac{5}{9})^{4-3}\\=4\times\frac{125}{729}\times\frac{4}{9}\\ =0.304832\\\approx0.305$

Thus, the probability that exactly three of the selected computers are desktops is 0.305.

(c)

Compute the probability that of the 4 computers selected at least 3 are desktops as follows:

$P(Y\geq 3)=1-P(Y$

Thus, the probability that at least three of the selected computers are desktops is 0.401.

6 0

3 years ago