Find the missing value

Find a formula for the sum of the first n even positive integers.

melisa1 [442]

So even postive integers are by defention in form 2k where k is a natural number so
let the sum of even integers to n=S
S=2(1+2+3+4+5+6+7+8+......+k-1+k
divide bith sides of equation 1 by 2
0.5S=1+2+3+4+5+...........+k-1+k
S=2(k+(k-1)+..............................+2+1)
divide both sides of equation 2 by 2
0.5S=k+k-1+..............................+2+1)
by adding both we will get
___________________________
S=(k+1)(k)
so the sum will be equal to
S=
${k}^{2} + k$
so let us test the equation
for the first 3 even number there sums will be
2+4+6=12
by our equation 3^2+3=12
gave us the same answer so our equation is correct

7 0

2 years ago

Use the formula A=2πrh to find the area of the curved surface of each of the cylinders below. (Express your answers correct to 1

shusha [124]

Answer:

here,

A=2×22÷7×17/2×21

A=22×17×3

A=1122 sq.cm

3 0

3 years ago

The mean per capita income is 16,127 dollars per annum with a variance of 682,276. What is the probability that the sample mean

MakcuM [25]

Answer:

0.60% probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

The standard deviation is the square root of the variance. So

$\mu = 16127, \sigma = \sqrt{682276} = 826, n = 476, s = \frac{826}{\sqrt{476}} = 37.86$

What is the probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

Either it differs by 104 or less dollars, or it differs by more than 104 dollars. The sum of the probabilities of these events is 100. I am going to find the probability that it differs by 104 or less dollars first.

Probability that it differs by 104 or less dollars first.

pvalue of Z when X = 16127 + 104 = 16231 subtracted by the pvalue of Z when X = 16127 - 104 = 16023. So

X = 16231

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{16231 - 16127}{37.86}$

$Z = 2.75$

$Z = 2.75$ has a pvalue of 0.9970

X = 16023

$Z = \frac{X - \mu}{s}$

$Z = \frac{16023 - 16127}{37.86}$

$Z = -2.75$

$Z = -2.75$ has a pvalue of 0.0030

0.9970 - 0.0030 = 0.9940

99.40% probability that it differs by 104 or less.

What is the probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

p + 99.40 = 100

p = 0.60

0.60% probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

7 0

3 years ago

I need help with this problem??

aalyn [17]

No I think it’s a 90 degree rotation because the notations above PQR indicates only 1 rotation...

4 0

3 years ago

Suppose that you are in charge of evaluating teacher performance at a large elementary school. One tool you have for this evalua

Strike441 [17]

Answer:

a) Standard error = 2

b) Range = (76.08, 83.92)

c) P=0.69

d) Smaller

e) Greater

Step-by-step explanation:

a) When we have a sample taken out of the population, the standard error of the mean is calculated as:

$\sigma_m=\dfrac{\sigma}{\sqrt{n}}=\dfrac{10}{\sqrt{25}}=\dfrac{10}{5}=2$

where n is te sample size (n=25) and σ is the population standard deviation (σ=10).

Then, the standard error of the classroom average score is 2.

b) The calculations for this range are the same that for the confidence interval, with the difference that we know the population mean.

The population standard deviation is know and is σ=10.

The population mean is M=80.

The sample size is N=25.

The standard error of the mean is σM=2.

The z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_M=1.96 \cdot 2=3.92$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 80-3.92=76.08\\\\UL=M+t \cdot s_M = 80+3.92=83.92$

The range that we expect the average classroom test score to fall 95% of the time is (76.08, 83.92).

c) We can calculate this by calculating the z-score of X=79.

$z=\dfrac{X-\mu}{\sigma}=\dfrac{79-80}{2}=\dfrac{-1}{2}=-0.5$

Then, the probability of getting a average score of 79 or higher is:

$P(X>79)=P(z>-0.5)=0.69146$

The approximate probability that a classroom will have an average test score of 79 or higher is 0.69.

d) If the sample is smaller, the standard error is bigger (as the square root of the sample size is in the denominator), so the spread of the probability distribution is more. This results then in a smaller probability for any range.

e) If the population standard deviation is smaller, the standard error for the sample (the classroom) become smaller too. This means that the values are more concentrated around the mean (less spread). This results in a higher probability for every range that include the mean.

6 0

3 years ago

Find the missing value ​

Find the missing value