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Alexandra [31]
3 years ago
15

. A large tree, b, removes 1.5 kg of pollution from the air each year. A small tree, s, removes 0.04 kg of pollution each year.

An urban forest has 1,650 large and small trees. Together, these trees remove 1,818 kg of pollution each year. Which system of equations could be used to find the number of large and small trees in the forest?
Mathematics
1 answer:
aivan3 [116]3 years ago
3 0

Answer:

The system of equations are:

s + b = 1650.... Equation 1

0.04s + 1.5b = 1818........ Equation 2

The number of

Small trees = 450 trees

The number of Big trees = 1200 trees

Step-by-step explanation:

From the above question

Small trees = s

Large trees = b

An urban forest has 1,650 large and small trees.

Hence,

s + b = 1650.... Equation 1

s = 1650 -b

A large tree, b, removes 1.5 kg of pollution from the air each year. A small tree, s, removes 0.04 kg of pollution each year. Together, these trees remove 1,818 kg of pollution each year.

Hence:

0.04s + 1.5b = 1818........ Equation 2

We substitute 1650 - b for s

0.04(1650 - b) + 1.5b = 1818

66 - 0.04b + 1.5b = 1818

- 0.04b + 1.5b = 1818 - 66

1.46b = 1752

b = 1752 ÷ 1.46

b = 1200 trees

Solving for x

1650 - b = s

s = 1650 - 1200

s = 450 trees

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Part 1. Please show work. Please assist me with these math problems. ​
Vitek1552 [10]

Answer: 1) 2, 3, 4, 9, 32, 279, 8896, 2481705, 22077238784

              2) 2,490,924

              3) Neither

<u>Step-by-step explanation:</u>

S_n=S_{n-2}\cdot (S_{n-1}-1)\quad \\\\S_1=2\quad given\\S_2=3\quad given\\S_3=S_1(S_2-1)\quad =2(3-1)\quad =4\\S_4=S_2(S_3-1)\quad =3(4-1)\quad =9\\S_5=S_3(S_4-1)\quad =4(9-1)\quad =32\\S_6=S_4(S_5-1)\quad =9(32-1)\quad =279\\S_7=S_5(S_6-1)\quad =32(279-1)\quad =8,896\\S_8=S_6(S_7-1)\quad =279(8896-1)\quad =2,481,705\\S_9=S_7(S_8-1)\quad =8896(2481705-1)\quad =22,077,238,784

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Neither

Not arithmetic because the difference between each of the terms is not the same.            

                        S₂ - S₁          S₃ - S₂           S₄ - S₃      

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Not geometric because the ratios between each of the terms is not the same.                

                       \dfrac{S_2}{S_1}=\dfrac{S_3}{S_2}\qquad \rightarrow \dfrac{3}{2}=\dfrac{4}{3}\qquad \rightarrow \text{cross multiply to get}: 9\neq 8

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4 0
3 years ago
An office manager has received a report from a consultant that includes a section on equipment replacement. the report indicates
wolverine [178]

Answer:

a) 22.663%

b) 44%

c) 38.3%

Explanation:

An office manager has received a report from a consultant that includes a section on equipment replacement. The report indicates that scanners have a service life that is normally distributed with a mean of 41 months and a standard deviation of 4 months. On the basis of this information, determine the percentage of scanners that can be expected to fail in the following time periods:

We solve the above question using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean = 41 months

σ is the population standard deviation = 4 months

a. Before 38 months of service

Before in z score score means less than 38 months

Hence,

z = 38 - 41/4

z = -0.75

Probability value from Z-Table:

P(x<38) = 0.22663

Converting to percentage = 0.22663 × 100

= 22.663%

b. Between 40 and 45 months of service

For x = 40 months

z = 40 - 41/4

z = -0.2

Probaility value from Z-Table:

P(x =40) = 0.40129

For x = 45

z = 45 - 41/4

z = 1

Probability value from Z-Table:

P(x = 45) = 0.84134

Between 40 and 45 months of service

= 0.84134 - 0.40129

= 0.44005

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= 0.44005 × 100

= 44.005%

= 44%

c. Wihin ± 2 months of the mean life

+ 2 months = 41 months + 2 months

= 43 month

- 2 months = 41 months - 2 months

= 39 months

For x = 43

z = 43 - 41 /4

z = 0.5

P-value from Z-Table:

P(x = 43) = 0.69146

For x = 39

z = 39 - 41/4

z = -2/4

z = -0.5

Probability value from Z-Table:

P(x = 39) = 0.30854

Within ± 2 months of the mean life

= 0.69146 - 0.30854

= 0.38292

= 38.3%

Learn more about z-score:

brainly.com/question/17436641

#SPJ4

4 0
2 years ago
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