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3 years ago

0.4074 rounded to the nearest thousand

Mathematics

2 answers:

luda_lava [24]3 years ago

8 0

0.4070 :) hope i helped you

Rzqust [24]3 years ago

8 0

Answer:

0.407

Step-by-step explanation:

Find the number in the thousandth place 7 and look one place to the right for the rounding digit 4 . Round up if this number is greater than or equal to 5 and round down if it is less than 5 .

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Can any one help me solve this ?

Neko [114]

The answer to your question is 12b^2
Because 3-12 is -9 pal :)

3 0

3 years ago

Scientific notation

tatiyna

Answer:

51,760

Step-by-step explanation:

Move the decimal place 4 tens places to the right (Positively)

51,760

I hope this helped!

7 0

3 years ago

Use the distributive property to simplify the expression. [a-4{2a-3(a-2)}]

bekas [8.4K]

A-8a+12-3a+6
The answer is 11a+18

4 0

3 years ago

Read 2 more answers

The Weibull distribution is widely used in statistical problems relating to aging of solid insulating materials subjected to agi

fenix001 [56]

Answer:

Step-by-step explanation:

Given that

$\alpha =2.5$ , $\beta =220$

The weibull distribution with parameters $\alpha \ \ and \ \ \beta$

where $\alpha =0 \ \ , \beta =0$

$F(x,\alpha ,\beta) \left|\begin{array}{cc}\frac{\alpha }{\beta } x^{\alpha-1e^-(x/\beta)^\alpha &x\geq 0\\0&x$

Then,

$F(x,\alpha ,\beta )=\left|\begin{array}{cc}0&x$

A) The probability that a specimen's lifetime is at most 250 is

$P(X\leq 250)=F(250,2.7,220)\\\\=1-e^-^(250/220)^{2.7}\\\\=1-0.2436\\\\=0.7564$

The probability that the specimen's life time is more than 300 is

$P(X>300)=1-P(X\leq 300)\\\\=1-F(300;2.7,220)\\\\=1-(1-e^-^{(300/220)^{2.7}$

$=e^-^{(300/220)^{2.7}$

= 0.0992

b)The probability of the specimen's lifetime is between 100 and 250

$P(100$

c) The value such that exactly 50% of all specimens have lifetimes exceeding that value is

$P(X>x)=0.50\\\\1-P(X$

x = 192.07

7 0

3 years ago

Why <img src="https://tex.z-dn.net/?f=%5Clim%20_%7Bx%5Cto%20-3%7D%5Cleft%28x%5E2-2x%2B4%5Cright%29" id="TexFormula1" title="\lim

shepuryov [24]

The quadratic is continuous over its entire domain, which is to say we can evaluate the limit by direct substitution:

$\displaystyle \lim_{x\to-3} (x^2 - 2x + 4) = (-3)^2 - 2(-3) + 4 = 9 + 6 + 4 = \boxed{19}$

You are mistaking (-3)² for -3². They are not the same number.

(-3)² = (-1 × 3)² = (-1)² × 3² = 1 × 9 = 9

-3² = -1 × 3² = -1 × 9 = -9

3 0

2 years ago