£13,819 to the nearest pound

Y = 2000-(1.05)

Aleonysh [2.5K]

Answer:

What is your initial value? 2000

Does this function represent exponential growth or exponential decay? Exponential growth

What is the rate of growth or rate of decay? 5%

7 0

2 years ago

Write each number in expanded form and using number names.

stellarik [79]

1. 20,000 +7,000 +500 +40 +9 -----WORD FORM :twenty-seven thousand,
five hundred forty-nine

2. Expanded Numbers Form:

 700,000 +90,000 +2,000 +0 +60 +5 
WORD FORM : seven hundred ninety-two thousand,
sixty-five

3 0

3 years ago

Pretty pls help me :3

lapo4ka [179]

Answer:

You can do 60 ÷ 6 = 10 groups.

Step-by-step explanation:

There are many ways to answer this and it's very simple.

6 and 10 is just one of the many factors of 60.

you can also do 60 ÷ 2 = 30 groups

3 0

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) $COV = 0.007255$

(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).

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