The letters that we use when are writing the exponent rules are called..

June had $240 more than Siti. Junr gave 25% of her money to Siti. Siti then gave 20% of her money to June. How much more money d

bulgar [2K]

Answer:

$144

Step-by-step explanation:

Let the money which Siti has be x

Therefore, the money June has will be x+240

First condition is that June gave 25% of her money to Siti, So this amount will be deducted from June's money and will be added in Siti's money.

Siti's money = x+ [25%(x+240)] June's money = x+240-[25%(x+240)]

= 1.25x + 60 = 0.75x + 180

Second condition is that Siti now gave 20% of her money to June.This amount will be deducted form Siti's money and added in June's money.

Siti's money = 1.25x+60-[20%(1.25x+60)]

= x+48

June's money = 0.75x +180+[20%(1.25x+60)]

= x+192

Difference between Sitit and June's money = x+192-x-48

= 144

Hence, June has $144 more money than Siti.

4 0

3 years ago

You work for a company that makes tiles for patios. A customer sent you the following picture of his patio. He said the patio is

Mumz [18]

Answer:3 by 1

Step-by-step explanation:

okay so I'll say the long side of a tile as x

the short side of a tile as y

look at the short side of the big patio first, it's made up of 2 big and 2 small

so 2x+2y=8 or x+y=8 so two added together is 8. So one possibility is 3 and 1 because (3+1)2 = 9

8 0

3 years ago

What is the median of the following number: 40, 49, 62, 56, 68, 39, 50, 61, 54, 44

Karo-lina-s [1.5K]

Answer:

52

Step-by-step explanation:

39, 40, 44, 49, 50, 54, 56, 61, 62, 68

39, 40, 44, 49, <u>50, 54,</u> 56, 61, 62, 68

$\frac{50 + 54}{2}$ = $\frac{104}{2}$ = 52

6 0

3 years ago

Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children. Let BBG ind

Gemiola [76]

Answer:

(a)

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

(b)

i.

$1\ girl = \{GBB, BBG, BGB\}$

$P(1\ girl) = 0.375$

ii.

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

$P(Atleast\ 2 \ girls) = 0.5$

iii.

$No\ girl = \{BBB\}$

$P(No\ girl) = 0.125$

Step-by-step explanation:

Given

$Children = 3$

$B = Boys$

$G = Girls$

Solving (a): List all possible elements using set-roster notation.

The possible elements are:

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

And the number of elements are:

$n(S) = 8$

Solving (bi) Exactly 1 girl

From the list of possible elements, we have:

$1\ girl = \{GBB, BBG, BGB\}$

And the number of the list is;

$n(1\ girl) = 3$

The probability is calculated as;

$P(1\ girl) = \frac{n(1\ girl)}{n(S)}$

$P(1\ girl) = \frac{3}{8}$

$P(1\ girl) = 0.375$

Solving (bi) At least 2 are girls

From the list of possible elements, we have:

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

And the number of the list is;

$n(Atleast\ 2 \ girls) = 4$

The probability is calculated as;

$P(Atleast\ 2 \ girls) = \frac{n(Atleast\ 2 \ girls)}{n(S)}$

$P(Atleast\ 2 \ girls) = \frac{4}{8}$

$P(Atleast\ 2 \ girls) = 0.5$

Solving (biii) No girl

From the list of possible elements, we have:

$No\ girl = \{BBB\}$

And the number of the list is;

$n(No\ girl) = 1$

The probability is calculated as;

$P(No\ girl) = \frac{n(No\ girl)}{n(S)}$

$P(No\ girl) = \frac{1}{8}$

$P(No\ girl) = 0.125$

7 0

3 years ago

Trapezoid ABCD has an area of 525 cm squared. If height AB = 21 cm and BC = 37 cm, what is the measure of AD?

Annette [7]

Answer:

AD = 13

Step-by-step explanation:

Area of trapezoid ABCD

$= \frac{1}{2} (AB + AD) \times AB \\ \\ 525= \frac{1}{2} (37+ AD) \times21\\ \\ 525 \times 2= (37+ AD) \times21 \\ \\ 1050 = 21 \times 37 + 21AD \\ \\ 1050 = 777+ 21AD \\ \\ 21AD = 1050 - 777 \\ \\ 21AD = 273 \\ \\ AD = \frac{273}{21} \\ \\ AD = 13$

8 0

3 years ago