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2 years ago

9

Create a scatterplot displaying the data in the table. Be sure to include a linear trend line. (2 points)

2 answers:

DENIUS [597]2 years ago

8 0

Answer:

Simply create a scatter plot then write in a line of best fit.

I'm not 100% sure

TEA [102]2 years ago

3 0

Answer:

I finished

Step-by-step explanation:

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Read the following two statements. Then, if possible, use the Law of Detachment to draw a conclusion.

jeyben [28]

<span>The hypothesis is : If Robbie wants to save money to buy a car. Conclusion : he must get a part-time job.

Robbie already started the new job yesteday. So, that makes the hypothesis true.

So, the conclusion would be : Robbie will save money to buy a car/ Robbie will buy a car.</span>

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2 years ago

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Apples are on sale for 5 for $3, how much does one apple cost?

evablogger [386]

The cost of one apple is I think $1.68

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2 years ago

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Which functions are decreasing? Select all answers that are correct.

monitta

Answer:

The last two functions are decreasing!

Step-by-step explanation:

You can easily tell the functions are decreasing if the function is going down from L to R. If it is going up, from L to R, then it is increasing!

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3 years ago

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Which table represents y as a function of x?

dezoksy [38]

Answer:

1 is the answer see attached

Step-by-step explanation:

1 is a function

2 is not a function as it is x+y

3 is a boundary function of how many parameters

4 is not a function as such parameters are too long however many of the decimals divide into each other and appear functional.

3 0

3 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

7 0

3 years ago