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Morgarella [4.7K]
3 years ago
8

Each year for 4 years, a farmer increased the number of trees in a certain orchard by of the number of trees in the orchard the

preceding year. If all of the trees thrived and there were 6,250 trees in the orchard at the end of the 4-year period, how many trees were in the orchard at the beginning of the 4-year period?
Mathematics
1 answer:
Neko [114]3 years ago
7 0

Answer:

The number of trees at the begging of the 4-year period was 2560.

Step-by-step explanation:

Let’s say that x is number of trees at the begging of the first year, we know that for four years the number of trees were incised by 1/4 of the number of trees of the preceding year, so at the end of the first year the number of trees wasx+\frac{1}{4} x=\frac{5}{4} x, and for the next three years we have that

                             Start                                          End

Second year     \frac{5}{4}x --------------   \frac{5}{4}x+\frac{1}{4}(\frac{5}{4}x) =\frac{5}{4}x+ \frac{5}{16}x=\frac{25}{16}x=(\frac{5}{4} )^{2}x

Third year    (\frac{5}{4} )^{2}x-------------(\frac{5}{4})^{2}x+\frac{1}{4}((\frac{5}{4})^{2}x) =(\frac{5}{4})^{2}x+\frac{5^{2} }{4^{3} } x=(\frac{5}{4})^{3}x

Fourth year (\frac{5}{4})^{3}x--------------(\frac{5}{4})^{3}x+\frac{1}{4}((\frac{5}{4})^{3}x) =(\frac{5}{4})^{3}x+\frac{5^{3} }{4^{4} } x=(\frac{5}{4})^{4}x.

So  the formula to calculate the number of trees in the fourth year  is  

(\frac{5}{4} )^{4} x, we know that all of the trees thrived and there were 6250 at the end of 4 year period, then  

6250=(\frac{5}{4} )^{4}x⇒x=\frac{6250*4^{4} }{5^{4} }= \frac{10*5^{4}*4^{4} }{5^{4} }=2560.

Therefore the number of trees at the begging of the 4-year period was 2560.  

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28,-(-73),(-65),(95),-(47) least to greatest
djyliett [7]

The larger the negative number, the least it is

Note, however, that two negatives = one positive

-65, -47, 28, -(-73) (or 73), 95 is your answer

hope this helps

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3 years ago
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How do you write "the difference between twenty and a number and four is six" as an algebraic expression?
kicyunya [14]
Using the variable x for the unknown number, 

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5 0
3 years ago
I need help with this question. check your solution please show WORK please. 5 = x + 8
PolarNik [594]

Answer:

x = -3

Step-by-step explanation:

5 = x + 8

You need to get x by Itself so subtract 8 to both sides.

5 - 8 = -3

so therefore x = - 3

4 0
3 years ago
Suppose that g(x) varies inversely with (x) and g(x) = 0.2 when x=0.1. What is g(x) when x=1.6?
mestny [16]
Since g(x) varies with x, therefore:
g(x) = k/x where k is a constant.

So, first we need to get k. We are given that g(x) = 0.2 when x = 0.1
Substitute with these values to get k as follows:
g(x) = k/x
0.2 = k/0.1
k = 0.2*0.1 = 0.02

Now, the equation became:
g(x) = 0.02 / x

We need to get the g(x) when x = 1.6
Therefore, we will substitute with x in the equation and calculate the corresponding g as follows:
g(x) = 0.02 / 1.6
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6 0
3 years ago
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A student carried out an experiment to determine the amount of vitamin C in a tablet sample. He performed 5 trials to produce th
ivolga24 [154]

Answer:

There is not enough evidence to support the claim that the amount of vitamin C in a tablet sample is different from 500 mg.

P-value = 0.166.

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{5}(490+502+505+495+492)\\\\\\M=\dfrac{2484}{5}\\\\\\M=496.8\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{4}((490-496.8)^2+(502-496.8)^2+(505-496.8)^2+(495-496.8)^2+(492-496.8)^2)}\\\\\\s=\sqrt{\dfrac{166.8}{4}}\\\\\\s=\sqrt{41.7}=6.5\\\\\\

Then, we can perform the hypothesis t-test for the mean.

The claim is that the amount of vitamin C in a tablet sample is different from 500 mg.

Then, the null and alternative hypothesis are:

H_0: \mu=500\\\\H_a:\mu< 500

The significance level is 0.05.

The sample has a size n=5.

The sample mean is M=496.8.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=6.5.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{6.5}{\sqrt{5}}=2.907

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{496.8-500}{2.907}=\dfrac{-3.2}{2.907}=-1.1

The degrees of freedom for this sample size are:

df=n-1=5-1=4

This test is a left-tailed test, with 4 degrees of freedom and t=-1.1, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=P(t

As the P-value (0.166) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the amount of vitamin C in a tablet sample is different from 500 mg.

4 0
3 years ago
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