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valentinak56 [21]
3 years ago
15

If a population of dolphins increases at a constant rate of 1.5% every year for 20 years what will be the total percentage incre

ase over the 20 years.
Mathematics
1 answer:
hichkok12 [17]3 years ago
6 0

Answer:

The population increased by 34.69% over 20 years.

Step-by-step explanation:

It is given that the population of dolphins increases at a constant rate of 1.5% every year for 20 years.

Formula for population increase:

P=a(1+r)^t

where, a is initial population, r is growth rate and t is time in years.

If the population of dolphins increases at a constant rate of 1.5% every year for 20 years, then the population after 20 years is

P=a(1+0.015)^{20}

P=a(1.015)^{20}

P=1.346855a

Where, a is the initial population.

The total percentage increase over the 20 years is

\% change=\frac{P-a}{a}\times 100

where, P is population after 20 years and a is initial amount.

\% change=\frac{1.346855a-a}{a}\times 100

\% change=\frac{0.346855a}{a}\times 100

\% change=0.346855\times 100

\% change=34.6855

\% change\approx 34.69

Therefore the population increased by 34.69% over 20 years.

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Answer:

B. 2106 pi mm³

Step-by-step explanation:

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2 years ago
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Which of the following options have the same value as 30\%30%30, percent of 818181?
Nataly_w [17]

Answer:

Option B is correct = 0.3 \times 81

Step-by-step explanation:

<u>The complete question is:</u> Which of the following options have the same value as 30% of 81?

Group of choices is:

(A) \frac{30}{100}\times 81 \times 100

(B) 0.3 \times 81

(C) 0.03 \times 81

(D) \frac{3}{10}\times 81 \times 10

(E) 30 \times 81

Now, the expression given to us is 30% of 81.

Simplifying the above expression we get;

   30% of 81  =  \frac{30}{100} \times 81

                     =  \frac{3}{10} \times 81  =  0.3 \times 81

Now, we will solve each of the given options and then see which option matches with our calculation.

Option (A) is given;

\frac{30}{100}\times 81 \times 100  =  30 \times 81

This doesn't match with our answer, so this option is not correct.

Option (B) is given;

0.3 \times 81  

<u><em>This matches with our answer, so this option is correct.</em></u>

Option (C) is given;

0.03 \times 81  

This doesn't match with our answer, so this option is not correct.

Option (D) is given;

\frac{3}{10}\times 81 \times 10  =  3 \times 81

This doesn't match with our answer, so this option is not correct.

Option (E) is given;

30 \times 81  

This doesn't match with our answer, so this option is not correct.

6 0
3 years ago
A physical fitness association is including the mile run in its secondary- school fitness test. the time for this event for boys
liberstina [14]

Answer:

<u>The probability that a randomly selected boy in school can run the mile in less than 348 seconds is 1.1%.</u>

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

μ of the time a group of boys run the mile in its secondary- school fitness test  = 440 seconds

σ of the time a group of boys run the mile in its secondary- school fitness test = 40 seconds

2. Find the probability that a randomly selected boy in school can run the mile in less than 348 seconds.

Let's find out the z-score, this way:

z-score = (348 - 440)/40

z-score = -92/40 = -2.3

Now let's find out the probability of z-score = -2.3, using the table:

p (-2.3) = 0.0107

p (-2.3) = 0.0107 * 100

p (-2.3) = 1.1% (rounding to the next tenth)

<u>The probability that a randomly selected boy in school can run the mile in less than 348 seconds is 1.1%.</u>

4 0
3 years ago
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A baseball player hit 65 home runs in a season. Of the 65 home​ runs, 19 went to right​ field, 23 went to right-center​ field, 9
olchik [2.2K]

Answer:

a) 29.23% probability that a randomly selected home run was hit to right field

b) 29.23% probability that a randomly selected home run was hit to right field, which is not lower than 5% nor it is higher than 95%. So it was not unusual for this player to hit a home run to right field.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes. It is said to be unusual if it is lower than 5% or higher than 95%.

(a) What is the probability that a randomly selected home run was hit to right field?

Desired outcomes:

19 home runs hit to right field

Total outcomes:

65 home runs

19/65 = 0.2923

29.23% probability that a randomly selected home run was hit to right field

(b) Was it unusual for this player to hit a home run to right field?

29.23% probability that a randomly selected home run was hit to right field, which is not lower than 5% nor it is higher than 95%. So it was not unusual for this player to hit a home run to right field.

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Katya went to a craft store to buy beads. The table shows her choices.
faltersainse [42]

Answer:

100 beads for $2.95

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