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3 years ago

A bag contains five red marbles and three green marbles.

1 answer:

7 0

1.3/8 because there is a total of 8 marbles but only 3 of them are green.
2. A 52 deck of cards has 4 or each card not including the Jokers so it would be 4/52.
3.30 red 90 total 30/20
4. all you have to do is 4*3=12
5. 4 of each card 2 red 2 black 2 red jacks 52 cards 2/52.
6.&7. coins aren't my cup of tea all i can say is maybe google it :P
8.2 pairs of blue socks 10 total pairs 2/10 or 1/5
sorry about 6 and 7 but i still hope this helped you

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For the purpose of purchasing new baseboard andâ€‹ carpet, complete parts a and b below.

Phantasy [73]

Answer:

(a)i. Area=72 square feet

ii. Perimeter =34 feet

(b) Baseboard - perimeter

Carpet - area

Step-by-step explanation:

The room is 9ft x 8ft, therefore it is rectangular shaped.

Length=9ft; Breadth=8ft

(a)i. Area of a Rectangle= Length X Breadth

Area of the room=8X9=72 square feet

ii. Perimeter of a Rectangle=2(Length+Breadth)

Perimeter of the room=2(8+9)=2 X 17 =34 feet

(b) A baseboard goes around the edge or boundary of the room, so it has to do with perimeter

A carpet covers the entire floor so it has to do with the area of the room.

3 0

3 years ago

I NEED THIS RN IMMEDIATELY ASAP PLS HELP !!! INSTRUCTIONS ARE ON THE FIRST PIC! (pls no links to other sites)

diamong [38]

Answer:

there are no instructions

Step-by-step explanation:

pls fix ur question

7 0

3 years ago

Read 2 more answers

How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?

BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

$\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}$

Where n = 8 and r = 0,1....8

When r = 0

$\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1$

When r = 1

$\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 2

$\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8$

When r = 3

$\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 4

$\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70$

When r = 5

$\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$