Helpp this is due tommorow c) Draw u + v + w.

(4x+5)(4x−5) 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? Pls answer A.S.A.P

WITCHER [35]

36

Step-by-step explanation:

360/2=180

180/5=36

7 0

3 years ago

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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

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