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Amanda [17]
3 years ago
6

A science class has 5 girls and 5 boys in the seventh grade and 3 girls and 5 boys in the eighth grade. The teacher randomly sel

ects a seventh grader and an eighth grader from the class for a competition. What is the probability that the students she selects are both boys? Write your answer as a fraction in simplest form.
Mathematics
2 answers:
kogti [31]3 years ago
7 0
Sample space of 7th grade (5 boys + 5 girls)

sample space of 8th grade (5 boys + 3 girls)


P(boy 7th grade) = 5/10 =1/2
P(boy 8th grade) = 5/8
Hence the probability of selecting a boy from 7th AND a boy from 8th is:

(1/2)x(5/8) ==> 5/16 ( It's a conditional probability)

alexgriva [62]3 years ago
3 0
So the ratios are 5:5 and 3:5 so the fractions are 5/10 (1/2) for 7th grade and 5/8 for 8th grade
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Which of the following is an example of a perfect square trinomial?
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c

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3 years ago
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Mariulka [41]

Answer:

Width: x + 3

(Length, width):

(5,4) (6,5) (7,6)

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x² + 7x + 12

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3 0
3 years ago
Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio
JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

Initial =1\ million

6\ years\ later = 500,000

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

\frac{dA}{dt} = kA

Where k represents the constant of proportionality

\frac{dA}{dt} = kA

Multiply both sides by dt/A

\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}

\frac{dA}{A}  = k\ dt

Integrate both sides

\int\ {\frac{dA}{A}  = \int\ {k\ dt}

ln\ A = kt + lnC

Make A, the subject

A = Ce^{kt}

t = 0\ when\ A =1\ million i.e. At initial

So, we have:

A = Ce^{kt}

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1000000 = Ce^{0}

1000000 = C*1

1000000 = C

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Substitute C =1000000 in A = Ce^{kt}

A = 1000000e^{kt}

To solve for k;

6\ years\ later = 500,000

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t = 6\ A = 500000

So:

500000= 1000000e^{k*6}

Divide both sides by 1000000

0.5= e^{k*6}

Take natural logarithm (ln) of both sides

ln(0.5) = ln(e^{k*6})

ln(0.5) = k*6

Solve for k

k = \frac{ln(0.5)}{6}

k = \frac{-0.693}{6}

k = -0.1155

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\frac{dA}{dt} = kA

Where

\frac{dA}{dt} = Rate

So, when

A = 600000

The rate is:

\frac{dA}{dt} = -0.1155 * 600000

\frac{dA}{dt} = -69300

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

7 0
2 years ago
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