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Licemer1 [7]
3 years ago
7

Mr. Henry has a class of 18 girls and 12 boys. He chose one student to come to the front and then chose a second student from th

ose still seated. What is the probability that both students chosen are girls?
Mathematics
2 answers:
Alekssandra [29.7K]3 years ago
6 0
Hi there!

The probability of choosing the first girl would be 18/30.

The probability of choosing the second girl would be 17/29.

The probability of choosing both girls would be 51/145 because 18/30 x 17/29 is 306/870 simplified to 51/145.

So the probability of choosing both girls would be 51/145.

Hope this helps :D
Oliga [24]3 years ago
4 0
18+12=number of students in his class (30)
probability that they are girls is 18(#of girls)/30(total students)
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guapka [62]

Answer:

Phrases 2 and 4.

Step-by-step explanation:

6.5x+1.5<21

The sum of 1.5 and the product of 6.5 and a number is no greater than 21.

The product of 6.5 and a number, when increased by 1.5 is at most 21.

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6 0
2 years ago
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pashok25 [27]

Answer:

Therefore the approximate area of this clock face is 415.3 feet².

Step-by-step explanation:

Given:

Clock is of Circular Shape,

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Radius =r =\dfrac{diameter}{2}=\dfrac{23}{2}=11.5\ feet

To Find :

Area of Clock Face = ?

Solution:

Area of Circle having Radius 'r' is given as,

\textrm{Area of Circle}=\pi r^{2}

Substituting the values we get

\textrm{Area of Clock Face}=3.14\times (11.5)^{2}=415.265=415.3\ feet^{2}

Therefore the approximate area of this clock face is 415.3 feet².

7 0
3 years ago
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Alexxandr [17]

Answer:

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Step-by-step explanation:

we have

(2x-3)^2

we know that

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In this problem

a=2x\\b=3

substitute in the formula

(2x-3)^2=(2x)^2-2(2x)(3)+(3)^2

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7 0
2 years ago
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Shkiper50 [21]

Answer:

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2) 6:22 pm

3) 6:42 pm

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6) 5:35 pm

7) 3:24am

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10) 9:20 am

6 0
3 years ago
Average Temperatures Suppose the temperature (degrees F) in a river at a point x meters downstream from a factory that is discha
mr Goodwill [35]

Answer:

Step-by-step explanation:

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a. [0, 10]

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T(0) = 160 - 0.05 × 0^2

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b. [10, 40]

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For x = 40

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c. [0, 40]

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T(0) = 160

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T(10) = 160 - 0.05 × 40^2

T(10) = 160 - 80 = 80

The average temperature

= (160 + 80)/2 = 120

6 0
2 years ago
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