In January of 2003(group 1), 1188 out of 1500 spots were bare ground (no vegetation). Find the sample proportion of bare ground spots.

Answers

Answer 1
Answer:

Answer:

n= 1500 represent the random sample selected

X= 1188 represent the number of pots that were bare ground (no vegetation

\hat p=(X)/(n)

And replacing we got:

\hat p=(1188)/(1500)= 0.792

So then the sample proportion of bare ground spots is 0.792 for this sample

Step-by-step explanation:

We have the following info given from the problem:

n= 1500 represent the random sample selected

X= 1188 represent the number of pots that were bare ground (no vegetation)

And for this case if we want to find the sample proportion of bare ground spots we can use this formula:

\hat p=(X)/(n)

And replacing we got:

\hat p=(1188)/(1500)= 0.792

So then the sample proportion of bare ground spots is 0.792 for this sample

Answer 2
Answer:

Final answer:

The sample proportion of bare ground spots is calculated by dividing the number of bare ground spots (successes) by the total number of spots (sample size). Using this method, the sample proportion for this scenario is 0.792 or 79.2%.

Explanation:

The student has asked about finding the sample proportion of bare ground spots. In Statistics, one defines a sample proportion as the number of successes in a sample divided by the number of observations in that sample. In this case, the 'success' is finding a spot of bare ground.

The sample size here is the total number of spots observed, which is 1500 (group 1) and the number of successes is the number of bare ground spots, which is 1188.

To calculate the sample proportion, one uses the formula: p = x/n, where 'x' is the number of successes and 'n' the sample size. So, for this scenario the sample proportion (p) would be: p = 1188/1500 = 0.792.

So, the sample proportion of bare ground spots is 0.792 or 79.2% when expressed as a percentage.

Learn more about Sample Proportion here:

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The function f(x)=101+25x2 is represented as a power series f(x)=∑n=0[infinity]cnxn. Find the first few coefficients in the power series. Some of the other problems only ask for the nonzero coefficients, but in this problem you are supposed to include those coefficents that are 0. Also, this problem is only asking for the coefficients of the power series. It does not want you to include the x's or the powers of x.

Answers

Answer:

\large a_0=101,\;a_1=0,\;a_2=25,\;a_n=0\;for\;n>2

Step-by-step explanation:

Since f(x) is a polynomial, it is a power series by itself with

\large a_0=101,\;a_1=0,\;a_2=25,\;a_n=0\;for\;n>2

On the other hand, the representation of a function as a power series around a given point is unique. This means that these are the only possible coefficients of f as a power series around 0.

Victor collects data on the price of a dozen eggs at 8 different stores.median: $ 1.55
Find the lower quartile and upper quartile of
the data set.
lower quartile: $
upper quartile: S
?
$1.39 $1.40 $1.44 $1.50 $1.60 $1.63 $1.65 $1.80

Answers

Answer:

Lower quartile: $1.42

Upper quartile: $1.64

Step-by-step explanation:

The median is the middle value when all data values are placed in order of size.

The ordered data set is:

$1.39 $1.40 $1.44 $1.50 $1.60 $1.63 $1.65 $1.80

There are 8 data values in the data set, so this is an even data set.

Therefore, the median is the mean of the middle two values:

\implies \sf Median\;(Q_2)=(\$1.50+\$1.60)/(2)=\$1.55

Place "||" in the middle of the data set to signify where the median is:

$1.39 $1.40 $1.44 $1.50 ║ $1.60 $1.63 $1.65 $1.80

The lower quartile (Q₁) is the median of the data points to the left of the median.  As there is an even number of data points to the left of the median, the lower quartile is the mean of the the middle two values:

\implies \sf Lower\;quartile\;(Q_1)=(\$1.40+\$1.44)/(2)=\$1.42

The upper quartile (Q₃) is the median of the data points to the right of the median.  As there is an even number of data points to the right of the median, the upper quartile is the mean of the the middle two values:

\implies \sf Upper \;quartile\;(Q_1)=(\$1.63+\$1.65)/(2)=\$1.64

Answer:

to find the lower quartile and upper quartile of the given dataset, we need to first arrange the data in ascending order:

$1.39, 1.40, 1.44, 1.50, 1.60, 1.63, 1.65, 1.80$

The median of the dataset is given as $1.55$. Since there are an even number of data points, the median is the average of the two middle values, which in this case are $1.50$ and $1.60$.

Now, we need to find the lower quartile and upper quartile. The lower quartile is the median of the lower half of the data set, and the upper quartile is the median of the upper half of the data set.

The lower half of the dataset is $1.39, 1.40, 1.44, 1.50$. The median of this half is the average of the middle two values, which are $1.40$ and $1.44$.

Therefore, the lower quartile is $1.42$.

The upper half of the dataset is $1.60, 1.63, 1.65, 1.80$. The median of this half is the average of the middle two values, which are $1.63$ and $1.65$.

Therefore, the upper quartile is $1.64$.

Hence, the lower quartile of the dataset is $1.42$ and the upper quartile is $1.64$.

What to do for the me part ​

Answers

ANSWER:

Me: $100,00
Maximum total debt: $20,000
Maximum monthly payments: $833.30

Olve the system of equations below.3x+4y=10
6x-2y=40

(6,-2)

(2,6)

(2,-6)

(-2,-6)

Answers

Answer:

The correct solutions are (6, -2).

Step-by-step explanation:

For the first equation, rearrange to make x the subject.

3x + 4y = 10

3x = 10 -4y

Divide the whole equation by 3 to isolate x:

3x = 10 -4y\n3x / 3 = (10)/(3) - (4)/(3)y\nx = (10)/(3) - (4)/(3)y\n

Now substitute this into the second equation:

6x - 2y = 40\n6((10)/(3) - (4)/(3)y) - 2y = 40\n6((10)/(3)) + 6((4)/(3))  - 2y = 40

20 - 8y - 2y = 40\n20 - 10y = 40

Subtract 20 from both sides:

20 - 10y = 40

20 - 10y - 20 = 40 - 20

-10y = 20

Divide both sides by 2:

-10y ÷ 10 = 20 ÷ 10

-y = 2 ∴ y = -2

Plug this value back into the first equation:

3x + 4y = 10

3x + 4(-2) = 10

3x + (-8) = 10

3x - 8 = 10

Add 8 to both sides:

3x - 8 + 8 = 10 + 8

3x = 18

Divide both sides by 3:

3x ÷ 3 = 18 ÷ 3

x = 6

Therefore, the correct solutions are (6, -2).

Hope this helps!

Find the Greatest common factor of the numbers using lists of factors

8,12

Answers

The answer is 4 bc you can multiply 4 with both 8 and 12. I’m sorry but I’m bad at explaining but the answer is 4 :(

A meteorologist is studying the speed at which thunderstorms travel. A sample of 10 storms are observed. The mean of the sample was 12.2 MPH and the standard deviation of the sample was 2.4. What is the 95% confidence interval for the true mean speed of thunderstorms?

Answers

Answer:

The 95% confidence interval for the true mean speed of thunderstorms is [10.712, 13.688].

Step-by-step explanation:

Given information:

Sample size = 10

Sample mean = 12.2 mph

Standard deviation = 2.4

Confidence interval = 95%

At confidence interval 95% then z-score is 1.96.

The 95% confidence interval for the true mean speed of thunderstorms is

CI=\overline{x}\pm z*(s)/(√(n))

Where, \overline{x} is sample mean, z* is z score at 95% confidence interval, s is standard deviation of sample and n is sample size.

CI=12.2\pm 1.96(2.4)/(√(10))

CI=12.2\pm 1.487535

CI=12.2\pm 1.488

CI=[12.2-1.488, 12.2+1.488]

CI=[10.712, 13.688]

Therefore the 95% confidence interval for the true mean speed of thunderstorms is [10.712, 13.688].