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777dan777 [17]
3 years ago
5

Farmer frank owns a piece of grassland. The grass grows at a constant rate and each cow eats at a constant rate. 12 cows can eat

10 acres of grass in 28 days. 21 cows can eat 30 acres of grass in 63 days. How many cows can eat 72 acres of grass in 126 days?
Mathematics
1 answer:
MrRa [10]3 years ago
3 0

Answer: There are 36 cows that can eaten 72 acres of grass in 126 days.

Step-by-step explanation:

Let x be the amount of grass in an acre initially.

Let y be the amount of grass grown in an acre per day.

Since we have given that

12 cows can eat 10 acres of grass in 28 days

12×28  = 10x + 10×28 y

336 = 10x + 280 y----------------------------(1)

Similarly,

21 cows can eat 30 acres of grass in 63 days.

21×63  = 30x + 30×63 y

1323 = 30x +1890 y ---------------------------(2)

So, by graphing method, we get the value of x and y

x=\dfrac{126}{5},\ y=\dfrac{3}{10}

So, we have,  72 acres of grass in 126 days, we need to find the number of cows.

So, it becomes,

c×126 = 72x + 72×126 y

126c=72\times \dfrac{126}{5}+72\times 126\times \dfrac{3}{10}\\\\126x=4536\\\\x=\dfrac{4536}{126}\\\\x=36

Hence, there are 36 cows that can eaten 72 acres of grass in 126 days.

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oksian1 [2.3K]
According to the Central Limit Theorem, the distribution of the sample means is approximately normal, with the mean equal to the population mean (1.4 flaws per square yard) and standard deviation given by:
\frac{\sigma}{ \sqrt{n} }=\frac{1.2}{ \sqrt{178} }=0.09
The z-score for 1.5 flaws per square yard is:
z=\frac{1.5-1.4}{0.09}=1.11
The cumulative probability for a z-score of 1.11 is 0.8665. Therefore the probability that the mean number of flaws exceeds 1.5 per square yard is
1 - 0.8665 = 0.1335.
6 0
3 years ago
The amount that two groups of students spent on snacks in one day is shown in the dot plots below. Which statements about the me
RUDIKE [14]

Answer:

  • The mean for Group A is less than the mean for Group B.
  • The median for Group A is less than the median for Group B.
  • The mode for Group A is less than the mode for Group B.

Step-by-step explanation:

First, we can find the measures of center for each group.

<u>Group A</u>

Mode: 1

Median: (1 + 2) / 2 = 3 / 2 = 1.5

Mean: (1 * 5 + 2 * 4 + 3) / 10 = (5 + 8 + 3) / 10 = 16 / 10 = 1.6

<u>Group B</u>

Mode: 3

Median: 92 + 3) / 2 = 5 / 2 = 2.5

Mean: (1 * 3 + 2 * 2 + 3 * 4 + 5) / 10 = (3 + 4 + 12 + 5) / 10 = 24 / 10 = 2.4

From here, we can see that...

  • The mean for Group A is less than the mean for Group B.
  • The median for Group A is less than the median for Group B.
  • The mode for Group A is less than the mode for Group B.

Hope this helps!

8 0
4 years ago
The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar busi
eimsori [14]

Answer:

t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131  

Rejection Zone: t_{calculated} or t_{calculated}>2.131

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts  and data given  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X=4.2 represent the sample mean  

s=1.4 represent the sample standard deviation  

n=16 represent the sample selected  

\alpha=0.05 significance level  

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:    

Null hypothesis:\mu = 3.6    

Alternative hypothesis:\mu \neq 3.6    

If we analyze the size for the sample is < 30 and we know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic  

We can replace in formula (1) the info given like this:    

t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714

Critical values

On this case since we have a bilateral test we need to critical values. We need to use the t distribution with df=n-1=16-1=15 degrees of freedom. The value for \alpha=0.05 and \alpha/2=0.025 so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:

"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"

And we got t_{crit}=\pm 2.131    

So the decision on this case would be:

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131  

Rejection Zone: t_{calculated} or t_{calculated}>2.131

Conclusion    

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

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3 years ago
The sum of 12.6, 31, and 5.4 is greater then 20
Dmitry [639]

Answer:

49>20

Step-by-step explanation:

12.6 + 31 + 5.4 = 49 >20

5 0
3 years ago
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