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3 years ago

14 Answer:

Mathematics

2 answers:

morpeh [17]3 years ago

5 0

Answer:

$9.92 per month

Step-by-step explanation:

A(t) 851.05 x .14^1/12 = 9.92891666667

Rounded to monetary dollars = $9.92

In banks this does get rounded up to £9.93 alternative end year and second year months but not usually the first months.

Savatey [412]3 years ago

3 0

Answer:

Around $12.05 or 12.0547

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Estimate 3 3/8 X 10 13/15

ivanzaharov [21]

Answer:

36.70

Hope this helps!

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2 years ago

Read 2 more answers

a driver won a recent race with a record average of about 200 mph. This record is about 2 3/11 times the average speed of the fi

Oksi-84 [34.3K]

Answer:
300
——
11

Step-by-step solution

200/2*3/11
100*3/11
300/11

300/11 or 27 3/11 or 27.27

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2 years ago

10x=-3-8x^2<br> Solve by factoring

GrogVix [38]

Answer:

10 x = -3 -8 x²

Step-by-step explanation:

6 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) $COV = 0.007255$

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

In a sale, normal prices are reduced by 20%. Andrew bought a saddle for his horse in the sale. The sale price of the saddle was

pshichka [43]

The formula for this is: (sale price) / (1 - percentage)
Percentage have to be in decimal form, so divide percentage by 100 to get the decimal form.

220/(1-.20) = $275

The original price was $275.

8 0

3 years ago