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lana66690 [7]
3 years ago
10

HELPPPP I NEED HELPPPPP

Mathematics
2 answers:
dolphi86 [110]3 years ago
6 0

Answer:

First one is cubed

second is quotient

third is product

fourth is three less

and fifth is three more

Step-by-step explanation:

Lunna [17]3 years ago
5 0

Answer:

x^3 = A variable cubed

b ÷ 3 = Quotient of some number and three

3a = Three more than some number

y - 3 = Three less than a variable

z + 3 = Product of an unknown value and three

Hope this Helps (✿◡‿◡)

Sry I took so long to do this.

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you own seven travel books and are taking five of them on vacation. in how many ways can you select the five books you will take
Ipatiy [6.2K]
It depends on how many options you have
You multiply the total number of books you have time 5 because you can choose 5
8 0
3 years ago
Please help if you can thank you
Stolb23 [73]

Answer:

c. -18.5x ≤ 407

Step-by-step explanation:

-18.5 x -22 = 407

-18.5 x 0 = 0

-18.5 x 6 = -111

-18.5 x 17 = -314.5

-18.5 x 60 = -1110

(All products are either less than or equal to 407)

4 0
3 years ago
P% of m PLEASE HELP IMMEDIATELY
Aleonysh [2.5K]

Answer:

pm/ 100 is answer.

Step-by-step explanation:

here, p% of m

=p/100× m

= pm/ 100 ...is answer.

<em><u>hope</u></em><em><u> </u></em><em><u>its</u></em><em><u> </u></em><em><u>what</u></em><em><u> </u></em><em><u>you are searching for</u></em><em><u>.</u></em><em><u>.</u></em>

4 0
3 years ago
Solve for k.<br><br><br><br> -11 = <br><br><br> k <br> 3 <br><br> + -12
Verdich [7]
1/3 or .33 because you move the 12 onto the other side getting 1=k3 and if you divide by three to get k alone, you get 1/3
6 0
3 years ago
Suppose you have a 6-face unfair dice with numbers 1,2,3,4,5,6 on each of its faces. If the probability distribution of throwing
kotegsom [21]

Answer:

D is correct

Step-by-step explanation:

Here, we want to select which of the options is correct.

The correct option is the option D

Since the die is unfair, we expect that the probability of each of the numbers turning up

will not be equal.

However, we should also expect that if we add the chances of all the numbers occurring together, then the total probability should be equal to 1. But this does not work in this case;

In this case, adding all the probabilities together, we have;

1/12 + 1/12 + 1/12 + 1/12 + 1/12 + 1/2

= 5(1/12) + 1/2 = 5/12 + 1/2 = 11/12

11/12 is not equal to 1 and thus the probability distribution cannot be correct

4 0
4 years ago
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