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2 years ago

A certain forest covers an area of

2 answers:

8 0

Okay. I'll use a formula that my math teacher taught me.

3400(1-0.07)^15
3400(0.93)^15
3400(0.33670086205)
1144.78293098 ( which rounds to about 1145 square km )

Ede4ka [16]2 years ago

6 0

Answer:

1144.78 Km^2

Step-by-step explanation:

If the forest covers an area of 3400 km^2 and it decreased by 7%, the new area covered after a year will be equal to 3400 less the product of 3400 and 7%

= 3400 - 7% * 3400

= 3400 (1 - 0.07)

After the second year, the area becomes

= 3400 (1 - 0.07) - 7% * 3400 (1 - 0.07)

= 3400 (1 - 0.07)(1 - 0.07)

= 3400 (1 - 0.07)²

Following this pattern, the area after 15 years

= 3400 (1 - 0.07)^15

= 1144.782931 km^2

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IXL 7th grade math could someone please help!

KIM [24]

Answer:

supplementary angles are two angles that when add equal 180 degrees.

The supplement of an angle is 180 - n.

Step-by-step explanation:

n = 17(180 - n)

n = 3060 - 17n

n + 17n = 3060

18n = 3060

n = 3060/18

n = 170

so the unknown angle (n) = 170 degrees and its supplement = (180 - 170 = 10)......= 10 degrees.

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2 years ago

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Which length is equivalent to 12 yards?

elena-s [515]

Answer:

Step-by-step explanation:

there are 3 ft in a yard, so to find how many yards you multiply 3 time 12, which equals 36.

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2 years ago

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HELP ASAP WILL MARK BRAINLY

vodomira [7]

Answer:

$\frac{61}{99}$

Step-by-step explanation:

let x = 0.616161 (etc.)

make a second equation because you multiply by 100:

100x = 61.616161 (etc.)

subtract from each other

100x = 61.616161

x = 0.616161

you get

99x = 61

solve for x

$\frac{61}{99}$

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3 years ago

I need help with this please

Harlamova29_29 [7]

-12y+14-9y=14
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2 years ago

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases: a. Central area 5 .

Flauer [41]

Answer:

a) "=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got $t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228$

b) "=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got $t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086$

c) "=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got $t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845$

d) "=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got $t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678$

e) "=T.INV(1-0.01,25)"

And we got $t_{\alpha}= 2.485$

f) "=T.INV(0.025,5)"

And we got $t_{\alpha}= -2.571$

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

We will use excel in order to find the critical values for this case

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases:

a. Central area =.95, df = 10

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have $\alpha/2=0.025$ .

We can use the following excel codes:

"=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got $t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228$

b. Central area =.95, df = 20

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have $\alpha/2=0.025$ .

We can use the following excel codes:

"=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got $t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086$

c. Central area =.99, df = 20

For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have $\alpha/2=0.005$ .

We can use the following excel codes:

"=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got $t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845$

d. Central area =.99, df = 50

For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have $\alpha/2=0.005$ .

We can use the following excel codes:

"=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got $t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678$

e. Upper-tail area =.01, df = 25

For this case we need on the right tail 0.01 of the area and on the left tail we will have 1-0.01 = 0.99 , that means $\alpha =0.01$

We can use the following excel code:

"=T.INV(1-0.01,25)"

And we got $t_{\alpha}= 2.485$

f. Lower-tail area =.025, df = 5

For this case we need on the left tail 0.025 of the area and on the right tail we will have 1-0.025 = 0.975 , that means $\alpha =0.025$

We can use the following excel code:

"=T.INV(0.025,5)"

And we got $t_{\alpha}= -2.571$

8 0

3 years ago