A little help please 5 + (-4) ____ -14 + (-7) A. = B. < C. >

The aquarium is shaped like a rectangular prism with length 20 in, width 18 in, and height 10 in.

Sonbull [250]

Answer: 10 minutes

Step-by-step explanation:

First you have to find out how big the tank is. You will have to you the formula Length x Width x Height. 20 x 18 x 10= 3,600 in. Then you would divide that by 360. 3,600/360= 10. It would have to take 10 minutes to empty the aquarium.

7 0

3 years ago

A box contains 1079 counters which are coloured either black, white or red.

aleksley [76]

The total number of red counters that are in the box is; 728 red counters

We are told that the box contains 1079 counters.

Let the number of white counters be x.

Let the number of red counters be y.

Let the number of black counters be z.

Thus;

z:x = 2:7

Thus; 2x/7 = z

Also, we are given;

x:y = 3:8

Thus;

8x/3 = y

Since total number in the box is 1079,then;

x + y + z = 1079

x + (8x/3) + (2x/7) = 1079

Multiply through by 21 to get;

21x + 56x + 6x = 22659

83x = 22659

x = 22659/83

x = 273

2x/7 = z

Thus; z = 2 × 273/7

z = 78

8x/3 = y

y = 8 × 273/3

y = 728

Thus,there are;

273 white counters

728 red counters

78 black counters.

Read more at; brainly.com/question/11712611

4 0

2 years ago

20 customers are eating at a local resturant. of the 20 customers 4 have enough to pay cash, 16 have enough to pay card and 3 ha

Levart [38]

Answer:

19/20

Step-by-step explanation:

20 is the total number of people (the denominator) and 19 is the amount of people able to pay card (the numerator)

4 0

4 years ago

Find the probability of being dealt a full house. (Round your answer to six decimal places.)?

vesna_86 [32]

I'm assuming a 5-card hand being dealt from a standard 52-card deck, and that there are no wild cards.

A full house is made up of a 3-of-a-kind and a 2-pair, both of different values since a 5-of-a-kind is impossible without wild cards.

Suppose we fix both card values, say aces and 2s. We get a full house if we are dealt 2 aces and 3 2s, or 3 aces and 2 2s.

The number of ways of drawing 2 aces and 3 2s is

$\dbinom42\dbinom43=24$

and the number of ways of drawing 3 aces and 2 2s is the same,

$\dbinom43\dbinom42=24$

so that for any two card values involved, there are 2*24 = 48 ways of getting a full house.

Now, count how many ways there are of doing this for any two choices of card value. Of 13 possible values, we are picking 2, so the total number of ways of getting a full house for any 2 values is

$2\dbinom{13}2\dbinom42\dbinom43=3744$