Ask question

Ask question

All categories

2 years ago

6

Drag each tile to the correct box. Not all tiles will be used. Consider the expression below. Place the steps required to determ

ine the sum of the two expressions in the correct order.

1 answer:

Alik [6]2 years ago

7 0

Answer:

$\frac{3x+6}{x^{2}-x-6 } +\frac{2x}{x^{2} +x-12}=\frac{5x+12}{x^{2} +x-12}$

Step-by-step explanation:

The given expression is

$\frac{3x+6}{x^{2}-x-6 } +\frac{2x}{x^{2} +x-12}$

First, we need to factor each part of the expression

$\frac{3(x+2)}{(x+2)(x-3)} +\frac{2x}{(x-3)(x+4)}$

Remember that quadractic expression are factored in two binomial factors. The first quadratic expression factors are about two numbers which product is 6 and which difference is one. The second quadratic expression is about two numbers which product is 12 and which difference is 1.

Now, we simplify equal expression at each fraction.

$\frac{3}{(x-3)} +\frac{2x}{(x-3)(x+4)}$

Then, we use the least common factor about the denominators to sum those fractions. In this case, the least common factor is $(x-3)(x+4)$ , because those are the factors present in the denominators.

Now, we divide each fraction by the least common factor, and then multiply the numeratos by its result.

$\frac{3(x+4)+2x}{(x-3)(x+4)}$

Finally, we multiply all products and sum like terms.

$\frac{3x+12+2x}{x^{2} +4x-3x-12}=\frac{5x+12}{x^{2} +x-12}$

Therefore, the sum of the initial expression is equal to

$\frac{5x+12}{x^{2} +x-12}$

You might be interested in

In box kernel density estimation, _____________________. a. None of the options b. The histogram is decentralized over several d

lilavasa [31]

Answer:

b. The histogram is decentralized over several data points.

Step-by-step explanation:

Kernel density estimators can be classified as non-parametric density estimators. The Kernel density estimators first smooth each data point into a density bump, then sum them up to obtain the final density estimated curve. A good histogram analysis skill is reqired to understand kernel density estimators.

8 0

2 years ago

Read 2 more answers

Only 35% of the drivers in a particular city wear seat belts. Suppose that 20 drivers are stopped at random what is the probabil

lesya692 [45]

Here we have a situation where the probability of a driver wearing seat belts is known and remains constant throughout the experiment of stopping 20 drivers.

The drivers stopped are assumed to be random and independent.

These conditions are suitable for modelling using he binomial distribution, where

where n=number of drivers stopped (sample size = 20)

x=number of drivers wearing seatbelts (4)

p=probability that a driver wears seatbelts (0.35), and

C(n,x)=binomial coefficient of x objects chosen from n = n!/(x!(n-x)!)

So the probability of finding 4 drivers wearing seatbelts out of a sample of 20

P(4;20;0.35)

=C(20,4)*(0.35)^4*(0.65)^16

= 4845*0.0150061*0.0010153

= 0.07382

6 0

2 years ago

At a used book store, Valentina purchased three books for $2.65 each. If she paid with a $20 bill, how much change did she recei

Oksana_A [137]

Answer:

12.05$

Step-by-step explanation: 2.65$ x 3 - 20$= 12.05$

7 0

2 years ago

An certain brand of upright freezer is available in three different rated capacities: 16 ft3, 18 ft3, and 20 ft3. Let X = the ra

ruslelena [56]

Answer:

$E(X)=16\cdot 0.3+18\cdot 0.1+20\cdot 0.6=18.6$

$E(X^2)=16^2\cdot 0.3+18^2\cdot 0.1+20^2\cdot 0.6=349.2$

$V(X) = E(X^2)-[E(X)]^2=349.2-(18.6)^2=3.24$

The expected price paid by the next customer to buy a freezer is $466

Step-by-step explanation:

From the information given we know the probability mass function (pmf) of random variable X.

$\left|\begin{array}{c|ccc}x&16&18&20\\p(x)&0.3&0.1&0.6\end{array}\right|$

<em>Point a:</em>

The Expected value or the mean value of X with set of possible values D, denoted by <em>E(X)</em> or <em>μ </em>is

$E(X) = $\sum_{x\in D} x\cdot p(x)$

Therefore

$E(X)=16\cdot 0.3+18\cdot 0.1+20\cdot 0.6=18.6$

If the random variable X has a set of possible values D and a probability mass function, then the expected value of any function h(X), denoted by <em>E[h(X)]</em> is computed by

$E[h(X)] = $\sum_{D} h(x)\cdot p(x)$

So $h(X) = X^2$ and

$E[h(X)] = $\sum_{D} h(x)\cdot p(x)\\E[X^2]=$\sum_{D}x^2\cdot p(x)\\ E(X^2)=16^2\cdot 0.3+18^2\cdot 0.1+20^2\cdot 0.6\\E(X^2)=349.2$

The variance of X, denoted by V(X), is

$V(X) = $\sum_{D}E[(X-\mu)^2]=E(X^2)-[E(X)]^2$

Therefore

$V(X) = E(X^2)-[E(X)]^2\\V(X)=349.2-(18.6)^2\\V(X)=3.24$

<em>Point b:</em>

We know that the price of a freezer having capacity X is 60X − 650, to find the expected price paid by the next customer to buy a freezer you need to:

From the rules of expected value this proposition is true:

$E(aX+b)=a\cdot E(x)+b$

We have a = 60, b = -650, and <em>E(X)</em> = 18.6. Therefore

The expected price paid by the next customer is

$60\cdot E(X)-650=60\cdot 18.6-650=466$

4 0

2 years ago

A statistics teacher was interested in the relationship between the number of days students waited to start a project and the sc

Fiesta28 [93]

Answer:

<em>A: For each increase in the number of procrastination days by 1, the predicted grade decreases by 3.64 points.</em>

Step-by-step explanation:

<u>The slope of a Regression Line</u>

A straight line can be represented in the slope-intercept form:

y = mx + b

Where m is the slope and b is the y-intercept.

The slope describes how fast and in what direction the graph goes when x changes values.

If m is positive, increments in x imply increments in y.

If m is negative, increments in x imply decrements in y.

The regression line is:

ŷ = –3.64x + 96.5

Where:

x = the number of procrastination days

ŷ = the predicted grade

We can say the slope is m=-3.64. This means that:

A: For each increase in the number of procrastination days by 1, the predicted grade decreases by 3.64 points.

5 0

2 years ago

Read 2 more answers