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Verizon [17]
3 years ago
13

Use the equation below to find a, if b = 10 and c= 8. a=32-b-c​

Mathematics
1 answer:
Mars2501 [29]3 years ago
5 0

Answer:

a=14

Step-by-step explanation:

a= 32- (10) - (8)

a= 32-10-8

a=14

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Tpy6a [65]

Answer:

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4 0
3 years ago
What is 9 + (-16)?...............................
elena-14-01-66 [18.8K]

Answer:

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Step-by-step explanation:

4 0
3 years ago
8. For circle A, the circumference is 31.4 ft. What is the area? Use 3.14 for pi
telo118 [61]

Answer:

\huge\boxed{\sf A = 78.5\ ft\²}

Step-by-step explanation:

Circumference = C = 31.4 ft

We know that,

<h3>C = 2πr</h3>

Where, r is the radius

31.4 = 2(3.14)r

31.4 = 6.28(r)

Divide 6.28 to both sides

31.4/6.28 = r

5 ft. = r

<h3>r = 5 ft.</h3>

Now,

<h3><u>Finding area:</u></h3>

A = πr²

A = (3.14)(5)²

A = (3.14)(25)

<h3>A = 78.5 ft²</h3>

\rule[225]{225}{2}

7 0
2 years ago
Read 2 more answers
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ANTONII [103]
So the angle A on the first diamond corresponds with angle Q, angle B with angle S, angle C with angle R, and angle D with angle P. So if angle D correspond (equals) angle P then x+34=97 and if angle R corresponds with C then 3y-13=83. From there just do some basic algebra to find the x and y values. 
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
3 years ago
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