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frez [133]
3 years ago
7

Bluefin tuna are large fish that can weigh 1,500 pounds and swim at speeds of 55 miles per hour. because they are used in sushi,

a prime fish can be worth over $30,000. as a result, the western atlantic bluefin tuna have been exploited, and their numbers have declined exponentially. their numbers in thousands from 1974 to 1991 can be modeled by f (x) = 230(0.881)x, where x is the year and x = 0 corresponds to 1974. in what year were there or will there be 50 thousand tuna?
Mathematics
1 answer:
shtirl [24]3 years ago
3 0

Given exponential function

f(x)= 230 (0.881)^x

Where x is the number of years

f(x) is the number of tuna in thousands

To find out when number of tuna is 50 thousand , we plug in 50 for f(x) and solve for x

50 = 230(0.881)^x

Divide by 230 on both sides

\frac{5}{23}  = (0.881)^x

Take ln on both sides

ln(\frac{5}{23})  = ln(0.881)^x

ln(\frac{5}{23}) = x ln(0.881)

Divide by ln(0.881) on both sides

x = \frac{ln(\frac{5}{23})}{ln(0.881)}

x= 12.04

Year = 1974 + 12.04 = 1986

so in 1986 there will be 50 thousand tuna.

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zubka84 [21]
A=6
B= -1/2
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Anvisha [2.4K]

Answer:

x = 3, y = -4

Step-by-step explanation by substitution:

Solve the following system:

{4 x + 3 y = 0 | (equation 1)

5 y + 53 = 11 x | (equation 2)

Express the system in standard form:

{4 x + 3 y = 0 | (equation 1)

-(11 x) + 5 y = -53 | (equation 2)

Swap equation 1 with equation 2:

{-(11 x) + 5 y = -53 | (equation 1)

4 x + 3 y = 0 | (equation 2)

Add 4/11 × (equation 1) to equation 2:

{-(11 x) + 5 y = -53 | (equation 1)

0 x+(53 y)/11 = -212/11 | (equation 2)

Multiply equation 2 by 11/53:

{-(11 x) + 5 y = -53 | (equation 1)

0 x+y = -4 | (equation 2)

Subtract 5 × (equation 2) from equation 1:

{-(11 x)+0 y = -33 | (equation 1)

0 x+y = -4 | (equation 2)

Divide equation 1 by -11:

{x+0 y = 3 | (equation 1)

0 x+y = -4 | (equation 2)

Collect results:

Answer: {x = 3 , y = -4

_____________________________________

Solve the following system:

{4 x + 3 y = 0

5 y + 53 = 11 x

Hint: | Choose an equation and a variable to solve for.

In the first equation, look to solve for x:

{4 x + 3 y = 0

5 y + 53 = 11 x

Hint: | Isolate terms with x to the left hand side.

Subtract 3 y from both sides:

{4 x = -3 y

5 y + 53 = 11 x

Hint: | Solve for x.

Divide both sides by 4:

{x = -(3 y)/4

5 y + 53 = 11 x

Hint: | Perform a substitution.

Substitute x = -(3 y)/4 into the second equation:

{x = -(3 y)/4

5 y + 53 = -(33 y)/4

Hint: | Choose an equation and a variable to solve for.

In the second equation, look to solve for y:

{x = -(3 y)/4

5 y + 53 = -(33 y)/4

Hint: | Isolate y to the left hand side.

Subtract 53 - (33 y)/4 from both sides:

{x = -(3 y)/4

(53 y)/4 = -53

Hint: | Solve for y.

Multiply both sides by 4/53:

{x = -(3 y)/4

y = -4

Hint: | Perform a back substitution.

Substitute y = -4 into the first equation:

Answer: {x = 3 , y = -4

7 0
3 years ago
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pickupchik [31]
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The numbers 60-67 represent the lower quartile (lowest 25%). 25% of 60 is 15 not 30. Therefore it is incorrect.
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3 years ago
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