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2 years ago

30 POINTS! Combine the like terms of the 2 questions below:

Mathematics

1 answer:

astraxan [27]2 years ago

5 0

Answer:

$\tt 1)\: \boxed{\tt 4y+10}$

$\tt 2)\:\boxed{\tt 0.5x^4+10.5}$

Step-by-step explanation:

1) y + 4 + 3(y + 2)

Expand:-

$\tt y+4+3y+6$

Combine like terms:-

$\tt y+3y+4+6$

$\boxed{\tt 4y+10}$

2) 0.5(x⁴ - 3) + 12

Expand:-

$\tt 0.5x^4-1.5+12$

Add/Subtract Numbers:-

$\boxed{\tt 0.5x^4+10.5}$

___________________

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HELP FAST!

mote1985 [20]

3(2x + 4) = -6
6x + 12 = -6
6x = -6 - 12
6x = - 18
x = -18/6
x = -3

3 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) $COV = 0.007255$

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).

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