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3 years ago

256 to the nearest 1,000

Mathematics

1 answer:

gavmur [86]3 years ago

6 0

Answer:

0 or 256.000.

Step-by-step explanation:

Why 0? If you round 256 to the nearest 1,000, it will give you 0 because as the rounding rule says is the digit is greater than 500 it rounds to up, but if the digit is less than 500 it rounds to down. But as 256 is less than a thousand and less than 500 it equals 0. If you put this title: rounding calculator, you can prove that my answer is correct.

Why 256.000? If you round 256 to the nearest thousands (3 decimals), it will give you 256.000. My explanation: because when rounding numbers that are less than needed it adds the decimal place from left to right. ( the answer is correct but the explanation may not be ).

Hope this helps.

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The number of women at a training workshops was two less than three times the number of men. If 50 people attended the workshops

Drupady [299]

Set up a system of equations first

W: number of women
M: number of men

50=w+m
3m-2=w

Solve in by plugging in one equation into the other

50=(3m-2)+m
50=4m-2
52=4m
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So if m=13 plug it bag into one of the two equations at the beginning

50=13+w
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3 years ago

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Allushta [10]

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

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well then, so since this equation has that slope therefore

$\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{-5}{8}} ~\hfill \stackrel{reciprocal}{\cfrac{8}{-5}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{8}{-5}\implies \cfrac{8}{5}}}$

so we're really looking for the equation of a line whose slope is 8/5 and runs through (10,10)

$(\stackrel{x_1}{10}~,~\stackrel{y_1}{10}) ~\hspace{10em} \stackrel{slope}{m}\implies \cfrac{8}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{10}=\stackrel{m}{\cfrac{8}{5}}(x-\stackrel{x_1}{10})$

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2 years ago

Each side of a square calendar is 19 inches long. What is the calendar's perimeter?

liberstina [14]

Answer:

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3 years ago

Read 2 more answers

PLs hElP gIvInG DUE TODAY 23 points

taurus [48]

Answer:

Answer down below!

Step-by-step explanation:

1) 10

2) 4

3) 6

----------------

1 ) we know that perimeter = all the sides combined.

12+12+10+10=44

(44 is the given perim.; so therefore if 12x2+10x2=44, 20 is the missing width.

2 ) we know that area = lw.

10x4= 40

44 is the given area, so therefore if 10x4=40, 4 is the base.

3 ) we know that area = lw.

we need to find the area of the square. 3 x 2 = 6. 6 is the area of the missing square or the part that needs to be covered.

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2 years ago

The bacterial strain Acinetobacter has been tested for its adhesion properties, which is believed to follow a normal distributio

pantera1 [17]

Answer:

a) Alternative hypothesis should be one sided. Because Null and Alternative hypotheses are:

$H_{0}$ : μ=2.66 dyne-cm.

$H_{a}$ : μ<2.66 dyne-cm.

b) the hypothesis that mean adhesion is at least 2.66 dyne-cm is true

Step-by-step explanation:

Let μ be the mean adhesion in dyne-cm.

Null and alternative hypotheses are:

$H_{0}$ : μ=2.66 dyne-cm.

$H_{a}$ : μ<2.66 dyne-cm.

First we need to calculate test statistic and then the p-value of it.

test statistic of sample mean can be calculated as follows:

t= $\frac{X-M}{\frac{s}{\sqrt{N} } }$ where

X is the sample mean

M is the mean adhesion assumed under null hypothesis (2.66 dyne-cm)

s is the standard deviation known (0.7 dyne-cm_2)

N is the sample size(5)

Sample mean is the average of 2.69, 5.76, 2.67, 1.62 and 4.12 dyne-cm, that is $\frac{2.69+5.76+2.67+1.62+4.12}{5}$ ≈ 3.37

using the numbers we get

t= $\frac{3.37-2.66}{\frac{0.7}{\sqrt{5} } }$ ≈ 2.27

The p-value is ≈ 0.043. Taking significance level as 0.05, we can conlude that sample proportion is significantly higher than 2.66 dyne-cm.

Thus, according to the sample the hypothesis that mean adhesion is at least 2.66 dyne-cm is true

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3 years ago

256 to the nearest 1,000 ​

256 to the nearest 1,000