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Ann [662]
2 years ago
8

Helpp this is due tommorowc) Draw u + v + w.

Mathematics
1 answer:
Y_Kistochka [10]2 years ago
8 0
That shi tuff gang. Don’t know how to answer dat
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(4x+5)(4x−5) <br>I'm stuck can someone please help me ​
Marizza181 [45]

Answer:6small2−25

Step-by-step explanation:

5 0
3 years ago
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The Art club had an election to select a president. 40% of the 80 members of the club voted in the election. How many members di
Ghella [55]

Answer:

48 member did not vote

Step-by-step explanation:

32 is 40% of 80

80-32=48

7 0
3 years ago
There are 5 cocoa pods in a bag.
Virty [35]

Answer:

416 g

Step-by-step explanation:

The mean is calculated as

mean = \frac{sum}{count} , then

\frac{sum}{5} = 398 ( multiply both sides by 5 )

sum of 5 pods = 1990 g

let x be the weight of the sixth pod , then

\frac{1990+x}{6} = 401 ( multiply both sides by 6 )

1990 + x = 2406 ( subtract 1990 from both sides )

x = 416

That is the weight of the sixth pod is 416 g

6 0
2 years ago
What's the measurement of x?<br><br> Pls answer A.S.A.P
WITCHER [35]

36

Step-by-step explanation:

360/2=180

180/5=36

7 0
3 years ago
Read 2 more answers
2. The quality assurance department inspects its production line. The product either fails or passes the inspection. Past experi
Oksana_A [137]

Answer:

(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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