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irinina [24]
2 years ago
14

An electronics company makes communication devices for military contracts. The company just completed two contracts. The navy co

ntract was 2,300 devices and took 25 workers two weeks ( 40 hours per week) to complete. The army contract was for 5,500 devices that were produced by 35 workers in three weeks. On which contract were the workers more productive?
Mathematics
1 answer:
adelina 88 [10]2 years ago
4 0

Answer:

Workers of army contract were more productive.

Step-by-step explanation:

To determine on which contract the workers were more productive compute the work done per hour by the workers for both contracts.

  • <u>Navy Contract</u>:

Number of devices = 2,300

Number of workers = 25

Number of weeks = 2 weeks

Number of hours per week = 40

The number of hours it took 25 workers to produce 2300 devices is =

=25\times2\times40\\=2000\ hours

Then number of devices produced in 1 hour is = \frac{2300}{2000}= 1.15\ devices\ per\ hour

  • <u>Army Contract</u>:

Number of devices = 5,500

Number of workers = 35

Number of weeks = 3 weeks

Number of hours per week = 40

The number of hours it took 35 workers to produce 5500 devices is ==35\times3\times40\\=4200\ hours

Then number of devices produced in 1 hour is = \frac{5500}{4200}= 1.31\ devices\ per\ hour

The workers for the army contract are more productive since they produce 1.31 devices per hour whereas the workers for navy contract produces 1.15 devices per hour.

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The concentration of particles in a suspension is 50 per mL. A 5 mL volume of the suspension is withdrawn. a. What is the probab
kolezko [41]

Answer:

(a) 0.6579

(b) 0.2961

(c) 0.3108

(d) 240

Step-by-step explanation:

The random variable <em>X</em> can be defined as the number of particles in a suspension.

The concentration of particles in a suspension is 50 per ml.

Then in 5 mL volume of the suspension the number of particles will be,

5 × 50 = 250.

The random variable <em>X</em> thus follows a Poisson distribution with parameter, <em>λ</em> = 250.

The Poisson distribution with parameter λ, can be approximated by the Normal distribution, when λ is large say λ > 10.  

The mean of the approximated distribution of X is:

μ = λ  = 250

The standard deviation of the approximated distribution of X is:

σ = √λ  = √250 = 15.8114

Thus, X\sim N(250, 250)

(a)

Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows:

P(235

                             =P(-0.95

Thus, the value of P (235 < <em>X</em> < 265) = 0.6579.

(b)

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

P(48

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Thus, the value of P(48.

(c)

A 10 mL sample is withdrawn.

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

P(48

                             =P(-0.40

Thus, the value of P(48.

(d)

Let the sample size be <em>n</em>.

P(48

                             0.95=P(-z

The value of <em>z</em> for this probability is,

<em>z</em> = 1.96

Compute the value of <em>n</em> as follows:

z=\frac{\bar X-\mu}{\sigma/\sqrt{n}}\\\\1.96=\frac{48-50}{15.8114/\sqrt{n}}\\\\n=[\frac{1.96\times 15.8114}{48-50}]^{2}\\\\n=240.1004\\\\n\approx 241

Thus, the sample selected must be of size 240.

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3 years ago
Two fractions negative whose product is 5/8
Andreyy89
You could do -1/-2 times -5/-4 which would be 5/8
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2 years ago
Solve the following proportion Start Fraction x over 2 End Fraction equals Start Fraction 8 over 9 End Fraction (1 point) 144 0.
daser333 [38]

Answer:

1.78

Step-by-step explanation:

\frac{x}{2} = \frac{8}{9}

To solve for x  we need to cross multiply it

We have = sign inbetween the fractions so we cross multiply it

9*x = 2* 8

9x = 16

Now we divide by 9 on both sides

x= \frac{16}{9}

so x= 1.7777777778

When we round to two decimal places then

x= 1.78


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3 years ago
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3 years ago
Need help with these 2 problems
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