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kenny6666 [7]
3 years ago
9

23:

Mathematics
2 answers:
irina [24]3 years ago
8 0

Answer:

(-\frac{3}{7}, 0) -- x intercept

(0,3) -- y intercept

Step-by-step explanation:

Given

y = 7x + 3

Required

Determine the x and y intercept

For x intercept.

Set y = 0

So, we have:

y = 7x + 3

0 = 7x + 3

Collect Like Terms

7x = -3

Make x the subject

x = -\frac{3}{7}

Hence, the x intercept is: (-\frac{3}{7}, 0)

For the y intercept;

Set x = 0

So, we have:

y = 7x + 3

y = 7*0 + 3

y = 0 + 3

y = 3

Hence, the y intercept is: (0,3)

RUDIKE [14]3 years ago
6 0

AnswerB

Step-by-step explanation:

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(a) E(X) = 950

(b) $ COV = 0.007255$

(c) P(X > 980) = 0.00001\\\\

Step-by-step explanation:

The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.

probability of failure = q = 0.05

probability of success = p = 1 - 0.05 = 0.95

number of trials = n = 1000

(a) What is the expected number of non-defective units?

The expected number of non-defective units is given by

E(X) = n \times p \\\\E(X) = 1000 \times 0.95 \\\\E(X) = 950

(b) what is the COV of the number of non-defective units?

The coefficient of variance is given by

$ COV = \frac{\sigma}{E(X)} $

Where the standard deviation is given by

\sigma = \sqrt{n \times p\times q} \\\\\sigma = \sqrt{1000 \times 0.95\times 0.05} \\\\\sigma = 6.892

So the coefficient of variance is

$ COV = \frac{6.892}{950} $

$ COV = 0.007255$

(c) What is the probability of having more than 980 non-defective units?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.

P(X > 980) = 1 - P(X < 980)\\\\P(X > 980) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

P(X > 980)  = 1 - P(Z < \frac{979.5 - 950}{6.892} )\\\\P(X > 980)  = 1 - P(Z < \frac{29.5}{6.892} )\\\\P(X > 980)  = 1 - P(Z < 4.28)\\\\

The z-score corresponding to 4.28 is 0.99999

P(X > 980) = 1 - 0.99999\\\\P(X > 980) = 0.00001\\\\

So it means that it is very unlikely that there will be more than 980 non-defective units.

8 0
3 years ago
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