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3 years ago

140,562 rounded to the nearest thousands

Mathematics

2 answers:

BartSMP [9]3 years ago

5 0

Answer:

It's 141,000

Step-by-step explanation:

because 4 or lower doesn't raise the rounding up it stay's the same if it is 4 or lower. But 5 or more raises it up so anything five or more you can round up. They taught me that in 4th grade but now i'm in 6th grade wich is so hard.

Ket [755]3 years ago

4 0

141,000 is your answer here

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I need the answers to 1 and 2 please help

Serggg [28]

Answer:

you need to write the article name or a picture of the article to get a correct answer.

Step-by-step explanation:

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3 years ago

I WILL CHOOSE BRAINLIEST as long as it LOOKS correct<br><br> Solve 3 ( x − 4 ) = 12 x

omeli [17]

Answer:

The answer is (-4/3)

Step-by-step explanation:

3(x − 4) = 12x

Multiply (x – 4) by 3 we get,

3(x – 4) = 12x

3x – 12 = 12x

3x – 12x – 12 + 12 = 12x – 12x + 12

-9x = 12

-9x/(-9) = -12/9

x = -4/3

Thus, The value of x is -4/3

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2 years ago

Read 2 more answers

1) Lithium isotope rations are important to medicine, the 6Li/7Li ratio in a standard reference material was measured several ti

uysha [10]

Answer:

1) $0.0826052-2.776\frac{0.000013424}{\sqrt{5}}=0.082588$

$0.0826052+2.776\frac{0.000013424}{\sqrt{5}}=0.0826219$

b) $ME= 2.776\frac{0.000013424}{\sqrt{5}}=0.0000166653$

And we want 2/3 of the margin of error so then would be: $2/3 ME = 0.00001111$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And on this case we have that ME =0.00001111016 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (2)

Replacing we got:

$n=(\frac{2.776(0.000013424)}{0.00001111})^2 =11.25 \approx 12$

So the answer for this case would be n=12 rounded up to the nearest integer

Step-by-step explanation:

Information given

0.082601, 0.082621, 0.082589, 0.082617, 0.082598

We can calculate the sample mean and deviation with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=0.0826052$ represent the sample mean

$\mu$ population mean

s=0.000013424 represent the sample standard deviation

n=5 represent the sample size

Part 1

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom, given by:

$df=n-1=5-1=4$

The Confidence level is 0.95 or 95%, and the significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ , the critical value would be using the t distribution with 4 degrees of freedom: $t_{\alpha/2}=2.776$

Now we have everything in order to replace into formula (1):

$0.0826052-2.776\frac{0.000013424}{\sqrt{5}}=0.082588$

$0.0826052+2.776\frac{0.000013424}{\sqrt{5}}=0.0826219$

Part 2

The original margin of error is given by:

$ME= 2.776\frac{0.000013424}{\sqrt{5}}=0.0000166653$

And we want 2/3 of the margin of error so then would be: $2/3 ME = 0.00001111$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And on this case we have that ME =0.00001111016 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (2)

Replacing we got:

$n=(\frac{2.776(0.000013424)}{0.00001111})^2 =11.25 \approx 12$

So the answer for this case would be n=12 rounded up to the nearest integer

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3 years ago

What is the decimal equivalent of 1/3

Vesna [10]

Answer:

Step-by-step explanation:

1/3 0.333 repeated 0.333 times 100 =33.333…%

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3 years ago

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Mariana [72]

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