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4 years ago

13

To find out how fast a tree grows, you can measure its trunk. A giant red oak's diameter was 248 inches in 1965. The tree's diam

eter had grown to 251 inches in 2005. Find the average rate of change in the diameter of the tree per year.

2 answers:

Semmy [17]4 years ago

6 0

The answer is 6.25 because you take 251 and divide it by 40 and you will end up with 6.25. The number 40 comes from the years 1965 to 2005. And the you just divide

shepuryov [24]4 years ago

5 0

The average rate a year is .7

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Ahoy there, matey! Our pirate ship had 5/6 of a treasure chest full of gold.

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Answer:

1 $\frac{5}{24}$ teasure chests of gold

Step-by-step explanation:

Since this is a problem of adding or subtracting fractions, you will have to do one or the other to solve. Since the pirates already had $\frac{5}{6}$ of a treasure chest of gold and stole another $\frac{3}{8}$ of a treasure chest to add to their collection, we need to add these two fractions:

$\frac{5}{6}+\frac{3}{8}$

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8: 8, 16, 24

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3 years ago

At the dealership where she works, Sade fulfilled 3/10 of her quarterly sales goal in January and another 1/10 of her sales goal

skelet666 [1.2K]

Hi There!

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Problem #1:

At the dealership where she works, Sade fulfilled 3/10 of her quarterly sales goal in January and another 1/10 of her sales goal in February. What fraction of her quarterly sales goal had Sade reached by the end of February?

3/10 + 1/10 = 4/10

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Problem #2:

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Problem #3:

For a class experiment, Vina's class weighed a log before and after subjecting it to termites. Before subjecting it to termites, the log weighed 7/10 of a pound. After the termites, the log weighed 3/10 of a pound. How much weight did the termites take from the log?

7/10 - 3/10 = 4/10

The termites took 4/10 pound away from the log.

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Hope This Helps :)

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3 years ago

A researcher compares the effectiveness of two different instructional methods for teaching physiology. A sample of 178 students

xxTIMURxx [149]

Answer:

critical value = 1.645

The 90% confidence interval = ( -22.62, -17.58)

Step-by-step explanation:

Given that:

the sample size $n_1$ = 178

the sample size $n_2$ = 226

the sample mean $\overline x_1$ = 54.4

the sample mean $\overline x_2$ = 74.5

population standard deviation $\sigma_1$ = 18.58

population standard deviation $\sigma_2$ = 9.52

level of significance ∝ = 1 - 0.90 = 0.10

The critical value for $Z_{\alpha/2} = Z _{0.10/2} = Z_{0.005}$ is 1.645

For the construction of our confidence interval, we use 90% since that is used to find the critical value.

∴

The margin of error = $Z \times\sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}}$

$1.645 \times\sqrt{\dfrac{18.58^2}{178} + \dfrac{9.52^2}{226}}$

$1.645 \times\sqrt{\dfrac{345.2164}{178} + \dfrac{90.6304}{226}}$

$1.645 \times\sqrt{2.34042}$

$\simeq$ 2.52

The lower limit = $( \overline x_1 - \overline x_2) - (M.O.E)$

= ( 54.4-74.5) - (2.52)

= -20.1 - 2.52

= -22.62

The upper limit = $( \overline x_1 - \overline x_2) + (M.O.E)$

= ( 54.4-74.5) + (2.52)

= -20.1 + 2.52

= -17.58

The 90% confidence interval = ( -22.62, -17.58)

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3 years ago