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1 year ago

Factor −5x2 + 10x. −5x(x + 2) 5(−x2 + 10x) 5x(−x + 2) x(5x + 10)

Mathematics

1 answer:

Hitman42 [59]1 year ago

4 0

${ \red{ \bold{option \: (c)}}}$

Step-by-step explanation:

${ \purple{ \tt{ { - 5x}^{2} + 10x}}}$

Take ${ \purple{ \tt{5x}}}$ as common. Then,

${ \purple{ \tt{5x( - x + 2)}}}$

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Step-by-step explanation:

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3 years ago

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Find the percentage of vacationers from<br> question 10 who spent between $1500<br> and $2000.

mars1129 [50]

Answer:

The percentage change of vacationers is 33.3 %

Step-by-step explanation:

Given as :

The old value of the vacationers = $ 1500

The new value of the vacationers = $ 2000

Let the percentage variation = x %

Or, x % increase = $\dfrac{\tyextrm new value - \textrm old value}{\textrm old value}$ × 100

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3 years ago

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).

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