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The stemplot displays the selling prices of homes for ABC Realty in April.

Alexxandr [17]

The stemplot displays the entire values for the selling prices of homes for ABC Reality in April.

The correct set of data for the stemplot is the option;

<u>$459,000, $227,000, $67,000, $123,000, $213,000, $482,000, $85,000, $181,000, $225,000, $92,000, $181,000, $287,000, $196,000, $276,000</u>

<h3>Method by which the above option is selected</h3>

The data in the stemplot are;

459,000, 482,000, 213,000, 225,000, 276,000, 277,000, 287,000, 123,000, 181,000, 181,000, 196,000, 67,000, 85,000, 92,000

Given that there are no values in the 300,000s, and that 181,000 appear twice, the option that corresponds to the above data is therefore;

$459,000, $277,000, $67,000, $123,000, $213,000, $482,000, $85,000, $181,000, $225,000, $92,000, $181,000, $287,000, $196,000, $276,000

Learn more about stemplot here:

brainly.com/question/10572086

6 0

1 year ago

First make a substitution and then use integration by parts to evaluate the integral. (Use C for the constant of integration.) x

e-lub [12.9K]

Answer:

$(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C$

Step-by-step explanation:

Ok, so we start by setting the integral up. The integral we need to solve is:

$\int x ln(5+x)dx$

so according to the instructions of the problem, we need to start by using some substitution. The substitution will be done as follows:

U=5+x

du=dx

x=U-5

so when substituting the integral will look like this:

$\int (U-5) ln(U)dU$

now we can go ahead and integrate by parts, remember the integration by parts formula looks like this:

$\int (pq')=pq-\int qp'$

so we must define p, q, p' and q':

p=ln U

$p'=\frac{1}{U}dU$

$q=\frac{U^{2}}{2}-5U$

q'=U-5

and now we plug these into the formula:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int \frac{\frac{U^{2}}{2}-5U}{U}dU$

Which simplifies to:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int (\frac{U}{2}-5)dU$

Which solves to:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\frac{U^{2}}{4}+5U+C$

so we can substitute U back, so we get:

$\int xln(x+5)dU=(\frac{(x+5)^{2}}{2}-5(x+5))ln(x+5)-\frac{(x+5)^{2}}{4}+5(x+5)+C$

and now we can simplify:

$\int xln(x+5)dU=(\frac{x^{2}}{2}+5x+\frac{25}{2}-25-5x)ln(5+x)-\frac{x^{2}+10x+25}{4}+25+5x+C$

$\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}-\frac{5x}{2}-\frac{25}{4}+25+5x+C$

$\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C$

notice how all the constants were combined into one big constant C.

7 0

2 years ago

NEED HELP ASAP PLEASE!! The provided diagram of triangle ABC will help you to prove that the base angles of an isosceles triangl

Lunna [17]

Answer: The proof is mentioned below.

Step-by-step explanation:

Here, Δ ABC is isosceles triangle.

Therefore, AB = BC

Prove: Δ ABO ≅ Δ ACO

In Δ ABO and Δ ACO,

∠ BAO ≅ ∠ CAO ( AO bisects ∠ BAC )

∠ AOB ≅ ∠ AOC ( AO is perpendicular to BC )

BO ≅ OC ( O is the mid point of BC)

Thus, By ASA postulate of congruence,

Δ ABO ≅ Δ ACO

Therefore, By CPCTC,

∠B ≅ ∠ C

Where ∠ B and ∠ C are the base angles of Δ ABC.

7 0

2 years ago

a student has $1500 to deposit in a savings account. what is the lowest rate that would allow the student to earn $95 in simple

nika2105 [10]

rate = interest / principal

rate = 95/ 1500 = 0.0633333

rate = 6.33%

8 0

2 years ago

Read 2 more answers

What is 362,000 rounded to the nearest ten thousand

Anon25 [30]

Answer:

360,000

Step-by-step explanation:

4 0

2 years ago

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