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Ksju [112]
4 years ago
9

hanna shops for socks that cost $2.99 for each pair and blouses that cost $12.99 each. let x represent the number of pairs of so

cks purchased, and let y represent the number of blouses purchased. which equation model the purchases she made with $43.92?
Mathematics
2 answers:
blsea [12.9K]4 years ago
6 0
In this item, we let x be the number of pairs of socks and y be the number of blouses that were purchased by Hanna. The total amount that is spent for the socks and blouses are 2.99x and 12.99y, respectively. The equation that would best describe the given scenario is,
                              2.99x + 12.99y = 43.92
AlexFokin [52]4 years ago
3 0

D) 2.99x + 12.99y = 43.92.

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Using cross products

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4 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)\\\\

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
A chemical factory has 2 cylindrical chemical tanks, one containing Chemical X and the other containing Chemical Y. The tank con
Mashutka [201]
The tank with Chemical X "takes up" a space of 25ft³.  Ordinarily we think of something "taking up" space as being area or surface area; however, area is a square measurement, and this is cubic; this must be volume.  The volume of the tank with Chemical X is 1.5 times the volume of the tank containing Chemical Y; setting this up in an equation we would have
25 = 1.5<em>V</em>
We would divide both sides by 1.5 to get the volume of the tank containing Chemical Y:
\frac{25}{1.5}=\frac{1.5V}{1.5} \\16 \frac{2}{3}=V
To find the volume of a cylinder, we find the base area and multiply by the height.  We know the volume and we know the base area, so our equation to find the height of the tank containing Chemical Y would look like:
16 \frac{2}{3}=3.2h \\ 16 \frac{2}{3}=3 \frac {2}{10}h
We would now divide both sides by 3 2/10:
\frac{16 \frac{2}{3}}{3 \frac{2}{10}}= \frac{3 \frac{2}{10}h}{3 \frac{2}{10}}
This is the same as:
\frac{\frac{50}{3}}{\frac{32}{10}}=h \\ \\ \frac{50}{3}* \frac{10}{32}=h \\ \\ \frac{500}{96}
So the height of the tank containing Chemical Y is 500/96 = 5 5/24 feet.
8 0
3 years ago
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