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murzikaleks [220]
4 years ago
6

trent, pam, and linda will work together to paint a poster that is 4 feet long Each student will paint an equal length of the po

ster how many feet of the poster length will each student paint?
Mathematics
1 answer:
Tasya [4]4 years ago
8 0
The whole length of the poster is 4 feet. There are three people who will paint an equal length, so this means that we need to divide the 4 feet of the poster between 3 people.

4 ÷ 3 <span>≈ 1.3

Each student will paint 1.3 feet, or 1 foot and 4 inches of the poster.</span>
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Given the diagram below find the value of x if the area is 21 aquare meters
Oksanka [162]

Answer:

The answer is <u><em>7</em></u>!

Step-by-step explanation:

(x-4) (x) = 21, x^2 - 4x = 21, x^2 - 4x - 21 = 0, (x - 7) (x + 3) = 0, x = 7 and -3; x can't be negative so the answer is 7!

x = 7 and x - 4 (other side) = 3

5 0
3 years ago
Omg help with my homework PLEASE ASAP. what donu graph and how do i suow my workkkkk whats the answer
Gnoma [55]

Answer:

you have to start with you b which in your problem is 5 so put a point at 5 on the y-axis on your graph. then with the M (the fraction numbers) you have to move either up or down depending if there is a negative number. so yours is 1/2 so then you would start at the 5 point and go up 1 and right 2 then you just keep repeating till you get a few points and can make a line then you just do the same thing for the other equation. start at 3 on the y-axis and then go up 3 and right 4 then just keep repeating till you get a few point and can create a line at the end the lines should cross each other like perpendicular lines

Step-by-step explanation:

hope this helps:)

7 0
2 years ago
3 ÷ 1/2 what is it?​
Alja [10]

3\div\dfrac{1}{2}=3\cdot2=6

8 0
3 years ago
Read 2 more answers
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
I know it’s a lot but 23 points and will make brainliest!!!<br> Please do 5-10
hodyreva [135]

Answer:

5, 20 - 52 = x

6, x - 27 = 5

7, 4 + x = 6

8, 14 - x = 29

9, x - 54 = 12

10, x + 25 = 40

Step-by-step explanation:

hope this is what u wanted.

brainiest please

5 0
3 years ago
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