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4 years ago

10

I give a medal if you can figure this out fast! A certain forest covers an area of 4200 km2 . Suppose that each year this area d

ecreases by 8.5% . What will the area be after 8 years?

2 answers:

Mars2501 [29]4 years ago

8 0

The answer would be 525 kilometers2

oee [108]4 years ago

7 0

<span>1541.213 square km would be correct

</span>

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A family paid $142.80 which included $6.80 in taxes for 4 play tickets each ticket costs the same amount how much did each perso

borishaifa [10]

$142.80-$6.80=$136 divided by 4 equals $34

8 0

3 years ago

If you have a demand function of Qd=48-9P and a supply function of Qs=-12+6P,

jenyasd209 [6]

Answer:

The equilibrium quantity is 26.4

Step-by-step explanation:

Given

$Q_d = 48 - 9P$

$Q_s = 12 + 6P$

Required

Determine the equilibrium quantity

First, we need to determine the equilibrium by equating Qd to Qs

i.e.

$Q_d = Q_s$

This gives:

$48 -9P = 12 + 6P$

Collect Like Terms

$-9P - 6P = 12 - 48$

$-15P = -36$

Solve for P

$P = \frac{-36}{-15}$

$P =2.4$

This is the equilibrium price.

Substitute 2.4 for P in any of the quantity functions to give the equilibrium quantity:

$Q_d = 48 - 9P$

$Q_d = 48 - 9 * 2.4$

$Q_d = 26.4$

<em>Hence, the equilibrium quantity is 26.4</em>

7 0

3 years ago

Plz help me I’m not good at math

grigory [225]

Answer:

second one

Step-by-step explanation:

am pretty sure

6 0

3 years ago

Read 2 more answers

I need help on 5 - 11

nalin [4]

5-11.

Positive 5 minus negative 11=

11-5=6

Now is 11 greater or 5, 11 is. And what is the integer of 11? A negative.

Therefore, the answer is -6.

4 0

2 years ago

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.2.a. If the distri

zalisa [80]

Answer:

a

$P(\= X \ge 51 ) =0.0062$

b

$P(\= X \ge 51 ) = 0$

Step-by-step explanation:

From the question we are told that

The mean value is $\mu = 50$

The standard deviation is $\sigma = 1.2$

Considering question a

The sample size is n = 9

Generally the standard error of the mean is mathematically represented as

$\sigma_x = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_x = \frac{ 1.2 }{\sqrt{9} }$

=> $\sigma_x = 0.4$

Generally the probability that the sample mean hardness for a random sample of 9 pins is at least 51 is mathematically represented as

$P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_{x}} \ge \frac{51 - 50 }{0.4 } )$

$\frac{\= X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ \= X )$

$P(\= X \ge 51 ) = P( Z \ge 2.5 )$

=> $P(\= X \ge 51 ) =1- P( Z < 2.5 )$

From the z table the area under the normal curve to the left corresponding to 2.5 is

$P( Z < 2.5 ) = 0.99379$

=> $P(\= X \ge 51 ) =1-0.99379$

=> $P(\= X \ge 51 ) =0.0062$

Considering question b

The sample size is n = 40

Generally the standard error of the mean is mathematically represented as

$\sigma_x = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_x = \frac{ 1.2 }{\sqrt{40} }$

=> $\sigma_x = 0.1897$

Generally the (approximate) probability that the sample mean hardness for a random sample of 40 pins is at least 51 is mathematically represented as

$P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_x} \ge \frac{51 - 50 }{0.1897 } )$

=> $P(\= X \ge 51 ) = P(Z \ge 5.2715 )$

=> $P(\= X \ge 51 ) = 1- P(Z < 5.2715 )$

From the z table the area under the normal curve to the left corresponding to 5.2715 and

=> $P(Z < 5.2715 ) = 1$

So

$P(\= X \ge 51 ) = 1- 1$

=> $P(\= X \ge 51 ) = 0$

5 0

3 years ago