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Verdich [7]
3 years ago
7

Scientists calculate that the average temperature of the earth is increasing at a constant rate of 0.10 degrees a year. This sit

uation can be modeled by what type of function?
Mathematics
1 answer:
Montano1993 [528]3 years ago
3 0

Answer:

exponential growth function

Step-by-step explanation:

When dealing with a constant rate of growth as a percentage form the only way to model such a situation would be with an exponential growth function. This is because the amount that it is growing by is changing every time period cycle because the total amount keeps growing since it is a percentage. The exponential growth function would be the following...

f(x) = a(1+r)^{x}

where

  • a = initial value
  • r = percentage of growth in decimal form
  • x = is the number of years
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1 times 108
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Solve the equation. Then check your solution. 5 /9 d = 9 /10 a. 31 / 90 c. 1/2 b. 1 31 /90 d. 50 /81
miskamm [114]

Answer:

d in (-oo:+oo)

(5/9)*d = 9/10 // - 9/10

(5/9)*d-(9/10) = 0

(5/9)*d-9/10 = 0

5/9*d-9/10 = 0 // + 9/10

5/9*d = 9/10 // : 5/9

d = 9/10/5/9

d = 81/50

d = 81/50

Step-by-step explanation:

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The value of x is 40 but how do I get that I need to show work ?
Butoxors [25]

Answer:

120 ÷ 3 = 40

Step-by-step explanation:

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3 years ago
Marta is buying a violin that costs $1,300. She makes a down payment of $500 and will pay the balance in 12 equal payments.
elixir [45]

Answer:

12 payemst of 66.66

Step-by-step explanation:

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7 0
3 years ago
A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime
Shalnov [3]

Answer:

We conclude that the true average percentage of organic matter in such soil is different from 3%.

Step-by-step explanation:

We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.

<em>Let </em>\mu<em> = true average percentage of organic matter in such soil</em>

SO, <u>Null Hypothesis</u>, H_0 : \mu = 3%   {means that the true average percentage of organic matter in such soil is equal to 3%}

<u>Alternate Hypothesis</u>, H_A : \mu \neq 3%   {means that the true average percentage of organic matter in such soil is different than 3%}

The test statistics that will be used here is <u>One-sample t test statistics</u> because we don't know about the population standard deviation;

                           T.S.  = \frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }  ~ t_n_-_1

where,  \bar X = sample mean amount of organic matter = 2.481%

              s = sample standard deviation = 1.616%

              n = sample of soil specimens = 30

So, <u><em>test statistics</em></u>  =  \frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }  ~ t_2_9

                               =  -1.759

<u></u>

<u>Now, P-value of the test statistics is given by;</u>

       P-value = P( t_2_9 > -1.759) = <u>0.046</u> or 4.6%

  • If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.
  • If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

<em>Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.</em>

Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.

3 0
3 years ago
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