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3 years ago

In a high-school, 135 freshmen were interviewed.

Mathematics

1 answer:

charle [14.2K]3 years ago

6 0

Answer:

$\dfrac{118}{135}$

Step-by-step explanation:

Total Number of Students =135

Number who took all three subjects=5

Number who took PE and BIO=10

Number who took PE and ENG=15

Number who took BIO and ENG=7

Therefore:

Number who took PE and BIO only=10-5=5
Number who took PE and ENG only=15-5=10
Number who took BIO and ENG only =7 -5=2

Total Number of students who took exactly two subjects=5+10+2=17

Therefore:

Number of students who did not take exactly two subjects=135-17=118

The probability that a randomly-chosen student from this group did not take exactly two subjects

$=\dfrac{\text{Number of students who did not take exactly two subjects}}{\text{Total Number of students }} \\=\dfrac{118}{135}$

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Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

5 0

3 years ago

Solve the problem shown above^^

malfutka [58]

Answer:

Step-by-step explanation:

Since the lengths of the shelves are given in cm and the answer is to be expressed in cm, then let's convert the length of the whole board from 2.5 m to cm. Since there are 100 cm in a m, then multiply 2.5 by 100 to get 250 cm. Now let's start with the expressions for the shelves. The lengths of all the shelves are based on the length of the first shelf. You can usually tell which object is the main one because it is mentioned the most number of times. The first shelf is mentioned 3 times so that is the main shelf that the measurements of all the other shelves are based upon. Let's call the first shelf x. The second shelf is 18 cm longer than twice the length of the first so the second shelf is 2x + 18.

The third shelf is 12 cm shorter than the first so the third shelf is x - 12.

The last shelf is 4 cm longer than the first shelf so the last shelf is x + 4. Since he needs to use all 250 cm of the board, we add all the shelves together and set them equal to 250 cm:

x + 2x + 18 + x - 12 + x + 4 = 250 and

5x + 10 = 250 and

5x = 240 so

x = 48. The first shelf is 48 cm long. But we need the length of the second shelf. The expression for the second shelf is 2x + 18, so 2(48) + 18 = 114 cm.

8 0

4 years ago

At a diving meet, Trudy scored 8.85 on her first dive and 9.35 on her second dive. What was her combined score?

Sever21 [200]

Trudy's combined score is 18.2

6 0

3 years ago

5. Find the sum of the first 35 terms of the arithmetic sequence when a = 5 and d = 4

Elza [17]

Answer:

The sum of the first 35 terms of the arithmetic sequence when a = 5 and d = 4 is 2555.

Step-by-step explanation: