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2 years ago

Can someone help me with this?!

Mathematics

1 answer:

NNADVOKAT [17]2 years ago

4 0

Answer:

$\sqrt{0/9}$

Step-by-step explanation:

Have a good day :)

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The interquartile range of a data set is 10 units. What does this represent?

Zepler [3.9K]

The interquartile range is from Q1 to Q3 and to get this you have to subtract Q2 by Q3. The 10 units represent that only 10 units are fit in the given range.

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2 years ago

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iragen [17]

Step-by-step explanation:

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2 years ago

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

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2 years ago

Perimeter 94 cm and length 39. What’s the width

ruslelena [56]

Answer:

Step-by-step explanation:

Take the length and multiply it by 2. then take your answer and multiply it by the perimeter. Then, take that answer and divide it by 2. Hope this helps :)

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3 years ago

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The function c(p) represents the total cost for a factory to produce p parts. c(p)=12.5p+4500 What does the value 4500 represent

deff fn [24]

Answer:

Step-by-step explanation:

c(p)=12.5p+4500

12.5p means that for each part it will cost $12.50, but the initial cost is $4500.

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3 years ago

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