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3 years ago

The sum of 6 consecutive integers is 270. What is the second number in the sequence?

1 answer:

3 0

N+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)=270
We use these terms to represent the first six consecutive integers in the sequence
Now we group like terms:
6n+15=270
Minus 15 from both sides of the equation
6n=255
n=42.5 which is not an integer
Is this a joke question?
Besides that, the 2nd number will be 43.5

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1 year ago

An organism measures 2.9 x 10-5 centimeters in diameter. What is this number in standard notation?

stepladder [879]

Answer: The answer is A. Hope this helps :)

Step-by-step explanation:

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James states that quadrilateral

mr_godi [17]

Using the distance formula,

$AD=\sqrt{(-1-3)^{2}+(-3-(-4))^{2}}=\sqrt{17}\\\\BC=\sqrt{(5-9)^{2}+(1-0)^{2}}=\sqrt{17} \\\\\therefore \overline{AD} \cong \overline{BC}$

$AB=\sqrt{(-1-5)^{2}+(-3-1)^{2}}=\sqrt{52}=2\sqrt{13}\\\\CD=\sqrt{(9-3)^{2}+(0-(-4))^{2}}=\sqrt{52}=2\sqrt{13}\\\\\therefore \overline{AB} \cong \overline{CD}$

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1 year ago

Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be

hodyreva [135]

Answer:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

And the sample deviation with the following formula:

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

And the sample deviation with the following formula:

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

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3 years ago

What does 19.2 x 29.5 equal? Step by step instructions please

slava [35]

Answer:

566.4

Step-by-step explanation:

do multipilication

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2 years ago

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