A salesman at J&R Computer World sold a laptop for $1500. His commission was $75. What was his commission rate?

Distribute the following expression: 9(2g + 5)

Kisachek [45]

9(2g + 5) = 18g + 45

answer is b) 18g + 45

5 0

3 years ago

Read 2 more answers

Find the sum of the first 8 terms in the geometric sequence: 1, 3, 9, 27,

Butoxors [25]

The sum of the first 8 items in the geometric sequence:1, 3, 9, 27, ...etc is 3280.

Step-by-step explanation:

The given is,

sequence: 1,3,9,21,.. etc

Step:1

The given is based on multiplication of 3 to previous number,

( x - previous number, next number = 3 × $x$ )

First term : 1

Second term : 1 × 3 = 3

Third term : 3 × 3 = 9

Fourth term : 9 × 3 =27

Fifth term : 27 × 3 = 81

Sixth term : 81 × 3 = 243

Seventh term : 243 × 3 = 729

Eighth term : 729 × 3 = 2187

Step:2

Sum of first 8 terms $= 1 + 3 + 9 + 27 + 81 + 243 + 729 + 2187$

= 3280

Sum of first 8 terms = 3280

Result:

The sum of the first 8 items in the geometric sequence:1, 3, 9, 27, ...etc is 3280.

7 0

3 years ago

The length of a parallelogram exceeds its breadth by 30 cm. If the perimeter of the parallelogram is 2 m 60 cm, find the length

Nuetrik [128]

$\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}$

$\longrightarrow \sf{2(1 + b) = 260}$

$\longrightarrow \sf{2(x+30+x) = 260}$

$\longrightarrow \sf{ 2x+30 = \dfrac{260}{20} }$

$\longrightarrow \sf{2x + 30 =130}$

$\longrightarrow \sf{2x = 130 - 30}$

$\longrightarrow \sf{2x = 100}$

$\longrightarrow \sf{x= \dfrac{100}{2} }$

• $\longrightarrow \sf{b= \underline{50cm}}$

$\longrightarrow \sf{= x + 30}$

$\longrightarrow \sf{= 50 + 30}$

• $\longrightarrow \sf{ \underline{80cm}}$

$\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}$

$\large\bm{Breadth= \: 50cm}$

$\large\bm{Length= \: 80cm}$

3 0

2 years ago

An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem

Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

Case 1: both balls are white.

At the beginning we have $b+w$ balls. We want to pick a white one, so we have a probability of $\frac{w}{b+w}$ of picking a white one.

If this happens, we're left with $w-1$ white balls and still $b$ black balls, for a total of $b+w-1$ balls. So, now, the probability of picking a white ball is

$\dfrac{w-1}{b+w-1}$

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

$\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}$

Case 2: both balls are black

The exact same logic leads to a probability of

$\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}$

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

$\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of $\frac{w}{b+w}$ of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability $\frac{b}{b+w}$ of picking a black ball with both picks.

This leads to an overall probability of

$\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}$

Of picking two balls of the same colour.

We want to prove that

$\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Expading all squares and products, this translates to

$\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}$

As you can see, this inequality comes in the form

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k}$

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y$

And this is our case, because in our case we have

$x=b^2+w^2$
$y=b^2+w^2+2bw$ so, y has an extra piece and it is larger
$k=b+w$ which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2