Ask question

Ask question

All categories

3 years ago

6

Kelly works in the clothing store where she earns a 20% commission on the clothes she sells. Last Saturday she sold $500 in merc

handise. what is her commission for Saturday?

1 answer:

krek1111 [17]3 years ago

3 0

Her commission in saturday is 20%*500 / 0.2*500=$100

You might be interested in

<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20-12%3D-4x" id="TexFormula1" title="x^{2} -12=-4x" alt="x^{2} -12=-4x" align="abs

Anit [1.1K]

Answer:

The answer for Factorization is (x+6)(x-2) = 0 or Solving for x is 2 and -6.

Step-by-step explanation:

x² - 12 = -4x

Factorisation :

x² + 4x - 12 = 0

x² - 2x + 6x - 12 = 0

x(x-2)x + 6(x-2) = 0

(x+6)(x-2) = 0

Solve for x :

(x+6)(x-2) = 0

x + 6 = 0

x = -6

x - 2 = 0

x = 2

3 0

3 years ago

Weary of the low turnout in student elections, a college administration decides to choose an SRS of three students to form an ad

-Dominant- [34]

Answer:

$P(ABC) = 0.110592$

$P(ABC^c) = 0.119808$

$P(AB^cC) = 0.119808$

$P(A^cBC) = 0.119808$

$P(AB^cC^c) = 0.129792$

$P(A^cBC^c) = 0.129792$

$P(A^cB^cC) = 0.129792$

$P(A^cB^cC^c) = 0.140608$

Step-by-step explanation:

Given

$P(A) = P(B) = P(C) = 48\%$

Convert the probability to decimal

$P(A) = P(B) = P(C) = 0.48$

Solving (a): P(ABC)

This is calculated as:

$P(ABC) = P(A) * P(B) * P(C)$

This gives:

$P(ABC) = 0.48*0.48*0.48$

$P(ABC) = 0.110592$

Solving (b): $P(ABC^c)$

This is calculated as:

$P(ABC^c) = P(A) * P(B) * P(C^c)$

In probability:

$P(C^c) = 1 - P(C)$

So, we have:

$P(ABC^c) = P(A) * P(B) * (1 - P(C))$

$P(ABC^c) = 0.48 * 0.48 * (1 - 0.48)$

$P(ABC^c) = 0.48 * 0.48 * 0.52$

$P(ABC^c) = 0.119808$

Solving (c): $P(AB^cC)$

This is calculated as:

$P(AB^cC) = P(A) * P(B^c) * P(C)$

$P(AB^cC) = P(A) * [1 - P(B)] * P(C)$

$P(AB^cC) = 0.48 * (1 - 0.48)* 0.48$

$P(AB^cC) = 0.48 * 0.52* 0.48$

$P(AB^cC) = 0.119808$

Solving (d): $P(A^cBC)$

This is calculated as:

$P(A^cBC) = P(A^c) * P(B) * P(C)$

$P(A^cBC) = [1-P(A)] *P(B) * P(C)$

$P(A^cBC) = (1 - 0.48)* 0.48 * 0.48$

$P(A^cBC) = 0.52* 0.48 * 0.48$

$P(A^cBC) = 0.119808$

Solving (e): $P(AB^cC^c)$

This is calculated as:

$P(AB^cC^c) = P(A) * P(B^c) * P(C^c)$

$P(AB^cC^c) = P(A) * [1-P(B)] * [1-P(C)]$

$P(AB^cC^c) = 0.48 * [1-0.48] * [1-0.48]$

$P(AB^cC^c) = 0.48 * 0.52*0.52$

$P(AB^cC^c) = 0.129792$

Solving (f): $P(A^cBC^c)$

This is calculated as:

$P(A^cBC^c) = P(A^c) * P(B) * P(C^c)$

$P(A^cBC^c) = [1-P(A)] * P(B) * [1-P(C)]$

$P(A^cBC^c) = [1-0.48] * 0.48 * [1-0.48]$

$P(A^cBC^c) = 0.52 * 0.48 * 0.52$

$P(A^cBC^c) = 0.129792$

Solving (g): $P(A^cB^cC)$

This is calculated as:

$P(A^cB^cC) = P(A^c) * P(B^c) * P(C)$

$P(A^cB^cC) = [1-P(A)] * [1-P(B)] * P(C)$

$P(A^cB^cC) = [1-0.48] * [1-0.48] * 0.48$

$P(A^cB^cC) = 0.52 * 0.52 * 0.48$

$P(A^cB^cC) = 0.129792$

Solving (h): $P(A^cB^cC^c)$

This is calculated as:

$P(A^cB^cC^c) = P(A^c) * P(B^c) * P(C^c)$

$P(A^cB^cC^c) = [1-P(A)] * [1-P(B)] * [1-P(C)]$

$P(A^cB^cC^c) = [1-0.48] * [1-0.48] * [1-0.48]$

$P(A^cB^cC^c) = 0.52*0.52*0.52$

$P(A^cB^cC^c) = 0.140608$

5 0

2 years ago

What is the additive inverse of 16 5/7?<br> Α -16<br> B. -16 7/5<br> C. -16 5/7<br> D. 16 7/5

topjm [15]

The answer is letter C

5 0

2 years ago

Read 2 more answers

The box plots show the high temperatures in january and march for denmark in degrees fahrenheit. box plots titled average daily

spin [16.1K]

From the information, the mean temperature for January is most likely lower than the mean temperature for March.

There are a couple of clues to this assumption. First, the max temp is lower for January. Second, the interquartile range is lower in March. With both of these factors, it is safe to assume the mean is lower.

8 0

2 years ago

Read 2 more answers

Find the value of <br> (16/81)^3/4

AlexFokin [52]

Answer:

1.92636657×10^-3

Step-by-step explanation:

This question looks quite awful coz it might be (16+81)^3/4 rather than being (16/81)^3/4....But, if it is (16/81)^3/4 the answer is 1.92636657×10^-3

hope you find it helpful....Thanks very much...

5 0

2 years ago