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3 years ago

5) How many thousands are there in 2 158 621?

Mathematics

2 answers:

never [62]3 years ago

6 0

Answer:

2,158,621 = 2 (i think)

Sergio [31]3 years ago

3 0

2

Blah blah blah blah

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What is the value of x?

zaharov [31]

Answer: around 47-49

Step-by-step explanation:

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3 years ago

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Set A of six numbers has a standard deviation of 3 and set B of four numbers has a standard deviation of 5. Both sets of numbers

Alenkinab [10]

Given:

$\sigma_A=3$

$n_A=6$

$\sigma_B=5$

$n_B=4$

$\overline{x}_A=\overline{x}_B$

To find:

The variance. of combined set.

Solution:

Formula for variance is

$\sigma^2=\dfrac{\sum (x_i-\overline{x})^2}{n}$ ...(i)

Using (i), we get

$\sigma_A^2=\dfrac{\sum (x_i-\overline{x}_A)^2}{n_A}$

$(3)^2=\dfrac{\sum (x_i-\overline{x}_A)^2}{6}$

$9=\dfrac{\sum (x_i-\overline{x}_A)^2}{6}$

$54=\sum (x_i-\overline{x}_A)^2$

Similarly,

$\sigma_B^2=\dfrac{\sum (x_i-\overline{x}_B)^2}{n_B}$

$(5)^2=\dfrac{\sum (x_i-\overline{x}_B)^2}{4}$

$25=\dfrac{\sum (x_i-\overline{x}_B)^2}{4}$

$100=\sum (x_i-\overline{x}_B)^2$

Now, after combining both sets, we get

$\sigma^2=\dfrac{\sum (x_i-\overline{x}_A)^2+\sum (x_i-\overline{x}_B)^2}{n_A+n_B}$

$\sigma^2=\dfrac{54+100}{6+4}$

$\sigma^2=\dfrac{154}{10}$

$\sigma^2=15.4$

Therefore, the variance of combined set is 15.4.

7 0

3 years ago

Can someone help me with this?

denpristay [2]

Answer:

rational, Natural, Integer, whole

Step-by-step explanation:

remember your perfect squares

the sqrt of 25 is a perfect square

sqrt of 25 is 5

5 is all except irrational

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3 years ago

Use the law of exponets to rewrite (5x3)^7

topjm [15]

Answer: 5^7 x 3^7
You could also write 15^7

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3 years ago

What is 100.360 rounded to the nearest whole number plz help

balu736 [363]

If you round 100.360 to the nearest whole number it'll be (100) if you see "100.500" i'll be 101 due to the decimals

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3 years ago

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5) How many thousands are there in 2 158 621?​

5) How many thousands are there in 2 158 621?