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3 years ago

Find the slope using the points (-1,-4) and (2,-2)

Mathematics

2 answers:

Lisa [10]3 years ago

6 0

Answer:. The slope is 2/3

Step-by-step explanation:

Slope is Rise over Run. The difference in y-values -2 -(-4) = 2. The difference in x-values 2 -(-1)=3 So it's 2 over 3 for the slope.

tresset_1 [31]3 years ago

3 0

The slope 2/3 ..........hehdhd

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Read 2 more answers

A manufacturing company uses an acceptance scheme on items from a production line before they are shipped. The plan is a two-st

pishuonlain [190]

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a) The probability that a box containing 3 defectives will be shipped is $51.74\%$

b) The probability that a box containing only 1 defective will be sent back for screening is $17.39\%$

Step-by-step explanation:

a) The first step is to count the number of total possible random sets of taking a sample size of 4 items over 23 items of the box, so $\left[\begin{array}{ccc}23\\4\end{array}\right] =23C4=8855$

The second step is to count the number of total possible random sets of taking a sample size of 4 items over 20 items of the box (discounting the 3 defectives) as the possible ways to succeed, so $\left[\begin{array}{ccc}20\\4\end{array}\right] =20C4=4845$

Finally we need to compute $\frac{\# ways\ to\ succeed}{\# random\ sets\ of \ 4} =\frac{4845}{8855}=0.5471=P(S)$ , therefore the probability that a box containing 3 defectives will be shipped is $P(S)=54.71\%$

a) The first step is to count the number of total possible random sets of taking a sample size of 4 items over 23 items of the box, so $\left[\begin{array}{ccc}23\\4\end{array}\right] =23C4=8855$

The second step is to count the number of total possible random sets of taking a sample size of 4 items over 22 items of the box (discounting the defective 1) as the possible ways to succeed, so $\left[\begin{array}{ccc}22\\4\end{array}\right] =22C4=7315$

Then we need to compute $\frac{\# ways\ to\ succeed}{\# random\ sets\ of \ 4} =\frac{7315}{8855}=0.8260=P(S)$ , therefore the probability that a box containing 1 defective will be shipped is $P(S)=82.60\%$

Finally the probability that a box containing only 1 defective will be sent back for screening will be $P(BS)=1-P(S)=1-0.8260=0.1739=17.39\%$

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3 years ago