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Anuta_ua [19.1K]
3 years ago
15

Survey: On free time 90 students like watching TV, 20 like watching TV but not reading, 80 like reading, 40 do not like watching

TV. Question: Percent of total student like both watching televisan and reading
Mathematics
1 answer:
Dvinal [7]3 years ago
5 0

Answer:

 50 %

Step-by-step explanation:

From the diagram attached,

Percentage of total students that like watching television = [(WnR)/μ]×100......... Equation 1

The number of students that like watching television and reading (WnR) = 70

The total number of students (μ) = 40+20+70+10

The total number of students (μ) = 140

Substitute these values into equation 1

Percentage of total students that like watching television = (70/140)×100

Percentage of total students that like watching television = 1/2(100)

Percentage of total students that like watching television = 50 %

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Sam delivers 167 newspapers every morning, he receives 5 cents per paper seven day a week, what is his weekly pay
steposvetlana [31]

Answer:

$58,45 or 5845 cents

Step-by-step explanation:

Weekly pay = 167 * 0.05 * 7 = $58.45 or 5845 cents

4 0
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HELP!! <br><br><br>Solve each equation.
Vikki [24]

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Step-by-step explanation:

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3 years ago
Please help me to solve this​
zzz [600]

Answer:

Below in bold.

Step-by-step explanation:

First find the height by use Pythagoras theorem on the right triangle:

h = sqrt (5^2 - 4^2)

= sqrt 9

= 3.

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= 1/2 * 12 * 10

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Total area = area of 2 triangles + area 0f 3 rectangles

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= 12 + 30 + 40 + 50

= 132 cm^2.

3 0
3 years ago
Find, correct to the nearest degree, the three angles of the triangle with the given vertices. A(1, 0, −1), B(4, −3, 0), C(1
zavuch27 [327]

Answer:

Angle at vertex  A : 59.2° B: 61.7° C : 59.1°

Step-by-step explanation:

1. First find the length of the vectors  

AB = B-A = (4-1, 3-0, 0-1) = (3, 3, -1)

AC = C-A = (1-1, 4-0, 3-1) = (0, 4, 2)

BC = C-B = (1-4, 4-3, 3-0) = (-3, 1, 3)

2. Find magnitude of the vectors

|AB| = √(3^2+3^2+〖(-1)〗^2 )=√19  

|AC| = √(0^2+4^2+2^2 )=2√5

|BC| = √(〖(-3)〗^2+1^2+3^2 )=√19

3. Find the angles between them

cos θ = (a.b)/(|a||b|) -----> θ = arc cos ((a.b)/(|a||b|))

AB, AC : θ =arc cos (AB.AC)/(|AB||AC|) = arc cos ( (3.0+3.4+ -1.2)/(√19  x 2√5)=  10/(2√95) ) =59.136, which is approximately 59.14°

AB, BC : θ =arc cos (AB.BC)/(|AB||BC|) = arc cos( (3.-3+3.1+ -1.3)/(√19  x √19)=  (-9)/19 ) = 118.27°

Because of the direction of BC is pointing relative to AB this is the angle outside the triangle, and we should find the supplementary angle.

180-118.27 = 61.73°

If we used –(BC) = CB in the formula (just negate the numerator) we would have gotten the correct angle on first try.

The 3rd angle should be what’s left after subtracting from 180°;

180-61.73-59.14 = 59.13 °

You can confirm using the formula again :

AC, BC : θ =arc cos (AC.BC)/(|AC||BC|) = arc cos( (0.-3+4.1+ 2.3)/(√19  x 2√5)=  10/(2√95) = 59.13°

Rounding everything up so they add up to 180 degrees we have Angle at vertex  A : 59.2° B: 61.7° C : 59.1°

7 0
3 years ago
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notka56 [123]

Answer:

1. 2(x+2)(x+5)

2. 8(x+4)^{2}

Step-by-step explanation:

<u>Problem #1:</u>

1. Find the GCF (Greatest Common Factor)

GCF = 2

2. Factor out the GCF and simplify.

2(x^{2} +7x+10)

3. Factor x^{2} +7x+10

<u>Which two numbers add up to 7 and multiply to 10?</u>

2 and 5

<u>Rewrite the expression using the above.</u>

(x+2)(x+5)

4. Done!

2(x+2)(x+5)

<u>Problem #2:</u>

1. Find the GCF (Greatest Common Factor)

GCF=8

2. Factor out the GCF and simplify.

8(x^{2} +8x+16)

3. Use the perfect square formula. a^{2} +2ab+b^{2} =(a+b)^{2}

a = x\\b=4

(x+4)^{2}

4. Done!

8(x+4)^{2}

8 0
2 years ago
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