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vichka [17]
4 years ago
11

What’s the algebraic expression to this please help !!!!

Mathematics
1 answer:
Wewaii [24]4 years ago
8 0

2 × c ÷ 7 ÷ 2. This better?


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Which number has both 2 and 5 as prime factors?
musickatia [10]

Answer:

Any multiple of 10

Step-by-step explanation:

A factor is defined as "a number or quantity that when multiplied with another produces a given number or expression." For example, the factors of 20 are 1, 2, 4, 5, 10, and 20.

When we multiply 2 and 5 together, we get 10. Multiplying this by anything else besides 0 will still yield something that has 2 and 5 as factors, for example 2 times 5 times 3 is 30. As this is just the same thing as multiplying 10 by 3, the answer is any multiple of 10.

4 0
3 years ago
Multiple the binomials (x+3)^2
maxonik [38]

Answer:

=x^2+6x+9

Step-by-step explanation:

\left(x+3\right)^2\\\mathrm{Apply\:Perfect\:Square\:Formula}:\quad \left(a+b\right)^2=a^2+2ab+b^2\\a=x,\:\:b=3\\=x^2+2x\cdot \:3+3^2\\\mathrm{Simplify}\:x^2+2x\cdot \:3+3^2:\quad x^2+6x+9\\x^2+2x\cdot \:3+3^2\\\mathrm{Multiply\:the\:num\\bers:}\:2\cdot \:3=6\\=x^2+6x+3^2\\3^2=9\\=x^2+6x+9

8 0
3 years ago
PLEASE HELP WITH MY MATH!!!!!!!!!!!!!!!!!
lozanna [386]

I think it would be c if it were a greater rate I don't know if i am right I'll check back with you guys on the answer

3 0
4 years ago
Read 2 more answers
Please help me with this
Citrus2011 [14]

I don't know the answer because of how zoomed in it is. But, you could make a cordinate grid and plot all of the numbers. Then, you would connect all of the lines and you will see that one line is missing in the parallelogram. You finish off that line yourself and write down the cordanite that you needed to complete the parallelogram. You will lastly put U=answer (the cordanite you got from finishing off the parallelogram).

5 0
3 years ago
Sample spaces For each of the following, list the sample space and tell whether you think the events are equally likely:
Schach [20]

Answer and explanation:

To find : List the sample space and tell whether you think the events are equally likely ?

Solution :

a) Toss 2 coins; record the order of heads and tails.

Let H is getting head and t is getting tail.

When two coins are tossed the sample space is {HH,HT,TH,TT}.

Total number of outcome = 4

As the outcome HT is different from TH. Each outcome is unique.

Events are equally likely since their probabilities \frac{1}{4} are same.

b) A family has 3 children; record the number of boys.

Let B denote boy and G denote girl.

If there are 3 children then the sample space is

{GGG,GGB,GBG,BGG,BBG,GBB,BGB,BBB}

The possible number of boys are 0,1,2 and 3.

Number of boys      Favorable outcome    Probability

           0                      GGG                        \frac{1}{8}

           1                    GGB,GBG,BGG          \frac{3}{8}

           2                   GBB,BGB,BBG           \frac{3}{8}

           3                       BBB                         \frac{1}{8}

Since the probabilities are not equal the events are not equally likely.

c)  Flip a coin until you get a head or 3 consecutive tails; record each flip.

Getting a head in a trial is dependent on the previous toss.

Similarly getting 3 consecutive tails also dependent on previous toss.

Hence, the probabilities cannot be equal and events cannot be equally likely.

d) Roll two dice; record the larger number

The sample space of rolling two dice is

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Now we form a table that the number of time each number occurs as maximum number then we find probability,

Highest number        Number of times         Probability

           1                                   1                     \frac{1}{36}

           2                                  3                    \frac{3}{36}

           3                                  5                    \frac{5}{36}

           4                                  7                    \frac{7}{36}

           5                                  9                    \frac{9}{36}

           6                                  11                    \frac{11}{36}

Since the probabilities are not the same the events are not equally likely.

4 0
4 years ago
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