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Angelina_Jolie [31]
2 years ago
6

C/2 - 3 + 6/d when c=14 d=3

Mathematics
1 answer:
Nataliya [291]2 years ago
4 0

Answer:

6

Step-by-step explanation:

14/2 - 3 + 6/3

Divide 14 by 2 to get 7.

7−3+6/3

Subtract 3 from 7 to get 4.

4+6/3

Divide 6 by 3 to get 2.

4+2

Add 4 and 2 to get 6.

6

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Branlet will be marked so help out all my gemstones
Jobisdone [24]

Answer:

B is the right answer.

Step-by-step explanation:

56 + 0.08x = 164

0.08x = 164 - 56

0.08x = 108

0.08x/0.08 = 108/0.08

x = 1,350

To confirm that:

56 + 0.08x

56 + 0.08(1,350)

56 + 108

= 164

164 is the answer.

6 0
3 years ago
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Evaluate the following expression when x = -5 and y = 25.
kykrilka [37]

Answer:

<u>The correct answer is A. 3,120.</u>

Step-by-step explanation:

Let's solve the expression for x = -5 and y = 25, this way:

(5|x| - y³)/x

Replacing x and y, we have:

(5 * 5 - 25³)/-5 (Absolute value of -5 is 5 in the numerator but we keep it without changes in the denominator)

(25 - 15,625)/-5

-15,600/-5

3,120

<u>The correct answer is A. 3,120.</u>

3 0
3 years ago
I will mark brainlist to whoever answers first
Nesterboy [21]

Answer:

its D

Step-by-step explanation:

i did the math ez

6 0
2 years ago
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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
3 years ago
Find the unknown factor.<br> 0= blank x (8 x 3)
algol [13]
I think the answer is 0

8*3=24

24 *0=0
3 0
2 years ago
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