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3 years ago

Help Please!! I'm confused about this... (you can explain or screenshot the graph and show where the points are)

Mathematics

1 answer:

saul85 [17]3 years ago

8 0

Your first step is to isolate y and put it into slope-intercept form.

Then graph

See screenshots below

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Human visual inspection of solder joints on printed circuit boards can be very subjective. Part of the problem stems from the nu

irina [24]

Answer:

(a) The probability that a selected joint was judged to be defective by neither of the two inspectors is 0.906.

(b) The probability that a selected joint was judged to be defective by inspector B but not by inspector A is 0.0213.

Step-by-step explanation:

The sample of joints randomly selected is, n = 10,000.

Number of joints judged defective by inspector A is, n (A) = 727.

The probability that a joint is judged defective by inspector A is:

$P(A)=\frac{n(A)}{n} =\frac{727}{10000} =0.0727$

Number of joints judged defective by inspector B is, n (B) = 756.

The probability that a joint is judged defective by inspector B is:

$P(B)=\frac{n(B)}{n} =\frac{756}{10000} =0.0756$

Number of joints judged defective by at least one of the inspectors is,

n (At least 1) = 940.

The probability that a joint is judged defective by at least one of the inspectors is:

$P(At\ least\ 1)=\frac{n(At\ least\ 1)}{n} =\frac{940}{10000}=0.094$

(a)

Compute the probability that a selected joint was judged to be defective by neither of the two inspectors as follows:

P (At least 1) = 1 - P (Less than 1)

= 1 - P (None)

P (None) = 1 - P (At least 1)

$=1-0.094\\=0.906$

Thus, the probability that a selected joint was judged to be defective by neither of the two inspectors is 0.906.

(b)

Compute the probability that a selected joint was judged to be defective by inspector B but not by inspector A as follows:

P (B but not A) = P (At least 1) - P (A)

$=0.094-0.0727\\=0.0213$

Thus, the probability that a selected joint was judged to be defective by inspector B but not by inspector A is 0.0213.

6 0

3 years ago

a hot air balloon holds 3,249 cubic meters kf helium. the density of the helium is 0.1785 kilograms per cubic meter. how many ki

Radda [10]

TO determine the mass of the helium in the balloon, we make use of the density of the gas that is given in the problem and the volume of helium that is present. Density is a property of a material which describes the mass of a material per unit volume. By multiplying the volume to density, we get the mass of the helium gas in the balloon. We do as follows:

Mass = Density x Volume
Mass = 0.1785 kg / m^3 x 3249 m^3
Mass = 579.9465 kg

Therefore, the mass of helium gas in the hot air balloon would be about 579.9 kg.

4 0

3 years ago

Need help with these math questions

PolarNik [594]

Answer:

31200 m

Step-by-step explanation:

120 times 60 equals 7200 and 200 times 120 equals 24000 and add the two up you get 31200

5 0

3 years ago

Express 3.023 in P form where p and q are integers and q= 0 D

AnnZ [28]