rim programs need a legally defensible retention schedule to reduce risk and legal liability. true or false?

Answer:

Explanation:

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Rosie Milojevic, Business Development at iComplied

Answered 3 years ago

For those who run or work in a business that handles food, you understand the importance of food safety and regulation compliance as it pertains to not only to certification and licensing of a company, but also the life or death of those who consume the products. Because food safety is such an important issue, we want to ensure that all companies who work in the production, preparation, or processing of food have the tools and information they need to ensure they are in full compliance 100% of the time. One slip in food safety compliance could cost someone their health or life, and this information spreads mistrust in the general public toward the company who sold the product, and also the entire product across companies throughout an entire country and beyond. So how does each company working with food ensure they comply with each and every food safety regulation? Through excellent auditing tools.

Auditing in Food Safety Must Change

Auditing in food safety compliance is essential in ensuring that all regulations are being complied with. Most auditing practices for major companies are severely out of date. Many rely on several different forms, stacks of paperwork, and data stored on multiple different computers and databases. This is incredibly inefficient, as it can be easy to miss something when comparing one sheet to the next, and paper can easily be lost or destroyed. It is also time consuming for companies to share audits over separate computers and databases, as it must be passed down from one employee to the next. Technology has granted all food safety compliance managers a simple solution. Auditing apps are the new wave solutions for companies who want all of their data stored on one cloud database that is instantly accessible with real-time data for all employees. This means that no one ever gets left out of the loop, and no new information will fall through the cracks.

what is the relative resolution (i.e. how many discrete levels are there) of an oscilloscope with 2^n, 8-bit D/A converter has 256 levels, 12-bit D/A converter has 4096 levels and 16-bit converter has 65536 levels.

In the system development and prototyping phases, oscilloscopes are of course the preferred tool for doing these measurements, but these are principally constrained by the vertical resolution of the scope's front-end. For instance, an 8-bit oscilloscope has a dynamic range of 256:1, therefore the theoretically lowest signal on a range of 1 volts is 3.9 millivolts (mV). Higher sensitivity and offset range are required to view millivolt level ripple signals on a 3.3 volt bus. Additionally, signal levels at the scope input will be muted when employing high attenuation probes to prevent circuit loading, making it difficult to measure them unless the instrument has a high resolution.

The ratio of the largest input signal an oscilloscope can handle to the smallest signal amplitude it can detect is known as oscilloscope vertical resolution. Resolution is often measured by the analog-to-digital converter's bit count (ADC). The resolution is equal to the amount of bits multiplied by two. The resolution of an 8-bit converter is therefore 28 or 256:1. The resolution of a 12-bit converter is 4096:1, which is 16 times higher than that of an 8-bit converter.

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Answer:

ill try to help you :)

Explanation:

Z≤ -4.852 ft, Maximum efficiency is η≅ 0.88 ≅ 88% is the maximum discharge possible

Solution

Given Data:-

D = 14.62in, N = 2134 rc/min, T=20°C. At T= 20°C ɣ=ρg= 62.35 lb/ft³, vapor pressure. Pv = 49.2 lb/ft².

The efficienies at each flow rate is computal by using formula

η = ρgθH / (550) (bhp)

→ As we can See the maximum efficiency point is at θ = 6ft³/s (close to 6ft³/s)

Maximum efficiency is η≅ 0.88 ≅ 88%

b) Given NPSHR = 16 ft,hg=22ft. Zactual. = 9ft  (below the sea surface)

To avoid cavitation NPSH < Pa - Pv/ρg - Z - hf

Z < Pa - Pv/ρg - hf

Z < 2116 - 49.2/62.35 - 16 - 22 [1 atm = 2116 lb/ft2]

Z≤ -4.852 ft

-> Keeping the pump 9 ft below the surface gives 4.148 ft of marign against cavitation.

Hence it is Sufficient to avoid cavitation.

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Assuming complementary inputs are available, the minimum number of transistors needed to realize a two input xor gate is 4.

If two inputs work well together in the production process, they are said to be complements. Since the inputs are used in fixed proportions, if the price of one input lowers, more of it will be sought after, increasing the demand for the other input.

In this study, a model of anticompetitive exclusive contracts with complimentary inputs is built. A downstream company converts a variety of complementary inputs into finished goods. When complementary input providers hold a dominant position in a specific input market, upstream competition price within that market benefits both the downstream firm and the complementary inputs providers by driving up complementary input prices.

Thus, even when entry that is socially optimal is permitted, the downstream firm is unable to generate higher profits. Therefore, even in the absence of scale economies, downstream competition, and relationship-specific investment, the inefficient incumbent provider can prevent socially efficient entrance by employing exclusive contracts.

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They are hosts on the same network, it is true. Subnet 255:248:0:0, IP 1.7.255.254, IP 1.2.3.4

You will most frequently encounter the subnet 255.255.255.0. As a result, if two addresses match in the first three sections (reading from left to right), and both addresses have the subnet 255.255.255.0, they are in the same subnet.

The network address is represented by the first 24 bits (the subnet mask's number of ones). The host address is represented by the final 8 bits, or the number of zeros in the subnet mask. You are given the addresses listed below: Address for the network: 11000000.10101000.01111011.00000000 (192.168.123.0)

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A movement that I am familiar is  a religious movement known as Heavenly conscious movement and the things i want them to gain additional knowledge about a performance on:

A movement is known to be one that make use of  planning as well as unpredictability. The movement is one that is given identity, leadership, as well as  coordination by one or more organizations.

Therefore, A movement that I am familiar is  a religious movement known as Heavenly conscious movement and the things i want them to gain additional knowledge about a performance on:

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Answer:

i) r_t = 0.5101 m

ii) m' = 106.73 kg/s

iii) R_s = 1.26

P = 2359.8 kW

iv) β2 = 55.63°

Explanation:

We are given;

Stagnation pressure; T_01 = 290 K

Inlet velocity; C1 = 145 m/s

Cp for air = 1005 kJ/(kg·K)

Mach number; M = 0.96

Ratio of specific heats; γ = 1.4

Stagnation pressure; P_01 = 1 bar

rotational speed; N = 5500 rpm

Work done factor; τ = 0.92

Isentropic effjciency; η = 0.9

Stagnation temperature rise; ΔT_s = 22 K

i) Formula for Stagnation temperature is given as;

T_01 = T1 + C1/(2Cp)

Thus,making T1 the subject, we havw;

T1 = T_01 - C1/(2Cp)

Plugging in the relevant values, we have;

T1 = 290 - (145/(2 × 1005))

T1 = 289.93 K

Formula for the mach number relative to the tip is given by;

M = V1/√(γRT1)

Where V1 is relative velocity at the tip and R is a gas constant with a value of 287 J/Kg.K

Thus;

V1 = M√(γRT1)

V1 = 0.96√(1.4 × 287 × 289.93)

V1 = 0.96 × 341.312

V1 = 327.66 m/s

Now, tip speed is gotten from the velocity triangle in the image attached by the formula;

U_t = √(V1² - C1²)

U_t = √(327.66² - 145²)

U_t = √86336.0756

U_t = 293.83 m/s

Now relationship between tip speed and tip radius is given by;

U_t = (2πN/60)r_t

Where r_t is tip radius.

Thus;

r_t = (60 × U_t)/(2πN)

r_t = (60 × 293.83)/(2π × 5500)

r_t = 0.5101 m

ii) Now mean radius from derivations is; r_m = 1.5h

While relationship between mean radius and tip radius is;

r_m = r_t - h/2

Thus;

1.5h = 0.5101 - 0.5h

1.5h + 0.5h = 0.5101

2h = 0.5101

h = 0.5101/2

h = 0.2551

So, r_m = 1.5 × 0.2551

r_m = 0.3827 m

Formula for the area is;

A = 2πr_m × h

A = 2π × 0.3827 × 0.2551

A = 0.6134 m²

Isentropic relationship between pressure and temperature gives;

P1 = P_01(T1/T_01)^(γ/(γ - 1))

P1 = 1(289.93/290)^(1.4/(1.4 - 1))

P1 = 0.9992 bar = 0.9992 × 10^(5) N/m²

Formula for density is;

ρ1 = P1/(RT1)

ρ1 = 0.9992 × 10^(5)/(287 × 289.93)

ρ1 = 1.2 kg/m³

Mass flow rate at compressor inlet is;

m' = ρ1 × A × C1

m' = 1.2 × 0.6134 × 145

m' = 106.73 kg/s

iii) stagnation pressure ratio is given as;

R_s = (1 + ηΔT_s/T_01)^(γ/(γ - 1))

R_s = (1 + (0.9 × 22/290))^(1.4/(1.4 - 1))

R_s = 1.26

Work is;

W = C_p × ΔT_s

W = 1005 × 22

W = 22110 J/Kg

Power is;

P = W × m'

P = 22110 × 106.73

P = 2359800.3 W

P = 2359.8 kW

iv) We want to find the rotor angle.

now;

Tan β1 = U_t/C1

tan β1 = 293.83/145

tan β1 = 2.0264

β1 = tan^(-1) 2.0264

β1 = 63.73°

Formula for Stagnation pressure rise is given by;

ΔT_s = (τ•U_t•C1/C_p) × tan(β1 - β2)

Plugging in the relevant values;

22 = (0.92 × 293.83 × 145/1005) × (tan 63.73 - tan β2)

(tan 63.73 - tan β2) = 0.5641

2.0264 - 0.5641 = tan β2

tan β2 = 1.4623

β2 = tan^(-1) 1.4623

β2 = 55.63°

Answer:

The answer will be B i hope this helps

Answer:

it is alba kk

Explanation:

rbeacuse wrong

The probability of throwing a 3 on a six-sided dice, where each side has a number between 1 and 6, is 1/6.

This is because there is one favorable outcome (rolling a 3) out of six possible outcomes (rolling a 1, 2, 3, 4, 5, or 6). A six-sided die has six equally likely outcomes when rolled. These outcomes include the numbers 1 through 6. Since there is only one 3 on the die, the probability of rolling a 3 is the number of ways to get a 3 (which is 1) divided by the total number of possible outcomes (which is 6). This gives us a probability of 1/6 or approximately 0.167. In other words, if we roll the die many times, we can expect to get a 3 about one-sixth of the time.

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Using the principle of mathematical induction, we can prove the largest term property of Big-Oh for sums of more than two functions. We already know that the property holds for a sum of two functions. Let's assume it holds for m functions, and we will prove that it holds for m+1 functions.

Let fi(n) be positive functions such that fi(n) = O(gi(n)) for all 1 ≤ i ≤ m+1. We want to show that ∑ fi(n) = O(max{g1(n), g2(n), ..., gm+1(n)}).

Base case (m=1): We have f1(n) + f2(n) = O(max{g1(n), g2(n)}), which is already established.

Inductive step: Assume the property holds for m functions, i.e., ∑ fi(n) = O(max{gi(n)}). Now consider the sum of m+1 functions, i.e., ∑ fi(n) + fm+1(n).

By the inductive hypothesis, ∑ fi(n) = O(max{gi(n)}) and fm+1(n) = O(gm+1(n)). Let's say that h1(n) = max{gi(n)} and h2(n) = gm+1(n). Then, ∑ fi(n) = O(h1(n)) and fm+1(n) = O(h2(n)). By the two-function largest term property, ∑ fi(n) + fm+1(n) = O(max{h1(n), h2(n)}). Since max{h1(n), h2(n)} = max{gi(n), gm+1(n)} = max{gi(n)}, the property holds for m+1 functions.

Thus, by induction, the largest term property of Big-Oh applies to sums of more than two functions.

Analyzing the worst-case time complexity of the Iterative Mod algorithm, the while loop runs until r < d, and in each iteration, r is reduced by d. The worst-case occurs when r starts with the maximum value (n) and is reduced by the minimum value (1) in each step. Therefore, the while loop will run n times. Since the operations inside the loop take constant time, the overall time complexity is O(n).

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Answer:

a. To solve this problem, we can use the binomial probability formula:

P(X=k) = (n choose k) * p^k * (1-p)^(n-k)

where:

- X is the number of out-of-state vehicles in a sample of n vehicles

- k is the number of out-of-state vehicles we're interested in (in this case, k=3)

- p is the probability that a given vehicle is from out of state (in this case, p=0.25)

- n is the sample size (in this case, n=9)

Plugging in the values, we get:

P(X=3) = (9 choose 3) * 0.25^3 * 0.75^6

       = 84 * 0.0039 * 0.1785

       = 0.5907%

Therefore, the probability that exactly three of the next 9 vehicles are from out of state is 0.5907%.

b. To find the expected number of out-of-state vehicles in the next hour, we can use the formula:

E(X) = n * p

where:

- X is the number of out-of-state vehicles in a sample of n vehicles

- p is the probability that a given vehicle is from out of state (in this case, p=0.25)

- n is the sample size (in this case, n=140)

Plugging in the values, we get:

E(X) = 140 * 0.25

    = 35

Therefore, the expected number of vehicles from out of state in the next hour is 35.

c. To find the probability of the number of vehicles varying between 2 standard deviations from the mean number of those passing through the check point, we need to find the mean and standard deviation of the number of out-of-state vehicles in a sample of 140 vehicles.

The mean is simply the expected value we found in part b:

mean = 35

The variance of a binomial distribution is:

Var(X) = n * p * (1-p)

where:

- X is the number of out-of-state vehicles in a sample of n vehicles

- p is the probability that a given vehicle is from out of state (in this case, p=0.25)

- n is the sample size (in this case, n=140)

Plugging in the values, we get:

Var(X) = 140 *

hope this helps :o

Answer:

Hello your question is incomplete attached below is the missing part

a) p1 = 454.83 kPa,  p2 = 283.359 Kpa , p3 = 536.423 kpa , p4 = 860.959kPa

b) W12 = 3.4 kJ, W23 = -3.5875 KJ, W34 = -4.0735 KJ, W41 = 3.5875 KJ

c) 5

Explanation:

Given data:

mass of air ( m ) = 1/10 kg

adiabatic index ( k ) = 1.4

temperature for isothermal expansion = 250K

rate of heat transfer ( Q12 ) = 3.4 KJ

temperature for Isothermal compression ( T4 ) = 300k

final volume ( V4 ) = 0.01m ^3

a)  Calculate the pressure, in Kpa, at each of the four principal states

from an ideal gas equation

P4V4 = mRT4 ( input values above )

hence P4 = 860.959kPa

attached below is the detailed solution

b) Calculate work done for each processes

attached below is the detailed solution

C) Calculate the coefficient of performance

attached below is detailed solution

Shear and moment diagrams for the loaded beam are shown below: {image}Let the two different forces of the beam be F1 = 16kN,

w1 = 3kN/m, and

w2 = 5kN/m. Therefore, the reactions at point A are

RA = 3.25kN and

MA = 3.25kNm. At the same time, the reactions at point B are

RB = 18.75kN and

MB = 78.25kNm. For x > 0, the value of V is equal to (RA - w1x) and for x > 6, the value of V is equal to (RB - w1x).

Therefore, the shear force diagram for the given loaded beam is shown below: {image}The moment diagram for the given loaded beam is shown below:

Answers to the given questions are as follows:

When x = 10,

V = -4.25 kN, 1kN ≤ Absolute Vmax ≤ 10kN,

Absolute Vmax = 13.5 kN. When

x = 10,

M = -10.5 kNm, 1kNm ≤ Absolute Mmax ≤ 10kNm,

Absolute Mmax = 78.25 kNm.

Thus, the shear and moment diagrams of the loaded beam have been drawn and the given questions have been answered as well.

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Answer:

A. A only

Explanation:

Airbags are safety features added to vehicles to prevent or reduce the shock impact during a collision.

Disconnecting the negative battery cable would prevent power from reaching the airbag system, but the airbag system has a capacitor that stores charges at initial incidents and would deploy the bags. After a collision, the airbag leads are shot off preventing the airbag system power.

Pain receptors adapt slowly or not at all because lack of adaptation is important to survival.

Pain receptors, also known as nociceptors, are specialized sensory receptors that respond to various types of painful stimuli. These receptors are designed to respond quickly and consistently to harmful stimuli in order to alert the body to potential danger and promote survival. Unlike other sensory receptors, such as mechanoreceptors, pain receptors do not adapt or become desensitized to stimuli over time. This lack of adaptation allows pain receptors to continue transmitting signals to the brain, even in the presence of ongoing or repeated harmful stimuli, which is important for the body to take necessary protective actions.

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A V-8 engine that posses a firing order of 18436572 with number 1 cylinder on compression, the cylinder which would be on the exhaust stroke is number: A. 6.

An injector can be defined as a mechanical device that is designed and developed for the introduction of fuel into the internal combustion system of an engine, especially under pressure.

In Engineering, there are two (2) main types of fuel injection system and these include the following:

A sequential fuel injection is typically designed and developed to spray fuel into the internal combustion system of an engine based on the firing order that is configured within it by the manufacturer.

As a general rule, one of the two-paired spark plugs within the internal combustion system of an engine is always designed to have a negative polarity while the other spark plug is designed to have a positive polarity always.

In this context, we can infer and logically deduce that a V-8 engine that posses a firing order of 18436572 with number 1 cylinder on compression, the cylinder which would be on the exhaust stroke is number 6.

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Complete Question:

The coil polarity in a waste-spark system is determined by the direction in which the coil is wound (left-hand rule for conventional current flow) and can't be changed. for example, if a v-8 engine has a firing order of 18436572 and number 1 cylinder is on compression, which cylinder will be on the exhaust stroke?

A. 6

B. 4

C. 8

D. 2

Answer:

15 000 000 Ohms

Explanation:

1 Mega Ohm = 1 000 000 Ohms

So,

15 Mega ohms =15 000 000 Ohms

Answer:

Induced draft is a mechanical draft created by air pulled through the boiler firebox by a blower located in the breaching after the boiler.

Explanation:

Answer:

Engineer A results will be more accurate

Explanation:

Analytical method is better than numerical method. Engineer A has used analytical method and therefore his results will be more accurate because he used simplified method. Engineer B has used software to solve the problem related to heat transfer his results will be approximate.

Answer:

D) Lower efficiency and lower effectiveness.

Explanation:

Given;

Two finned surfaces with long fins which are identical,

with difference in the convection heat transfer coefficient,

The first finned surface has a higher convection heat transfer coefficient, but gives the same heat rate as the second, which will make it (first finned surface) to have lower efficiency and lower effectiveness than the second finned surface.

Therefore, the correct option is "(D) Lower efficiency and lower effectiveness"

The time required for this system to come to rest is equal to 9.87 seconds.

Given the following data:

In order to calculate the time required for this system to come to rest, we would have to determine the moment of inertia for gears A and C.

Mathematically, the moment of inertia for a gear can be calculated by using this formula:

I = mr²

Where:

For gear A, we have:

I = mr²

I = 0.675 × 0.04²

I = 0.675 × 0.0016

I = 1.08 × 10⁻³ kg·m².

For gear C, we have:

I = mr²

I = 3.6 × 0.1²

I = 3.6 × 0.01

I = 0.036 kg·m².

Next, we would convert the initial angular velocity of gear C in revolution per minutes (rpm) to radian per seconds (rad/s) as follows:

ωc₁ = 2000 × 2π/60

ωc₁ = 4000π/60

ωc₁ = 209.44 rad/s.

Also, the initial angular velocity of gears A and B is given by:

ωA₁ = ωB₁ = rc/rA × (ωc₁)

ωA₁ = ωB₁ = 0.15/0.06 × (209.44)

ωA₁ = ωB₁ = 2.5 × (209.44)

ωA₁ = ωB₁ = 523.60 rad/s.

Taking the moment about A, we have:

I_A·ωA₁ + rA∫F_{AC}dt - M(f)_A·t = 0

Substituting the given parameters into the formula, we have;

(1.08 × 10⁻³)·(523.60) + 0.06∫F_{AC}dt - 0.15t = 0

0.15t - 0.06∫F_{AC}dt = 0.56549    ......equation 1.

Similarly, the moment about B is given by:

0.15t - 0.06∫F_{BC}dt = 0.56549    ......equation 2.

Note: Let x = ∫F_{BC}dt + ∫F_{AC}dt

Adding eqn. 1 & eqn. 2, we have:

0.3t - 0.06x = (0.56549) × 2

0.3t - 0.06x = 1.13098      ......equation 3.

Taking the moment about A, we have:

Ic·ωc₁ - rC∫F_{AC}dt - rC∫F_{BC}dt - Mc(f)_A·t = 0

0.036(209.44) - 0.3t - 0.15(∫F_{BC}dt + ∫F_{AC}dt) = 0

0.3t + 0.15x = 7.5398      ......equation 4.

Solving eqn. 3 and eqn. 4 simultaneously, we have:

x = 30.5 Ns.

Time, t = 9.87 seconds.

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Answer:

Email etiquette is important

Explanation:

It is important to understand how to use correct email etiquette because it helps you communicate more clearly. It also makes you seem a bit more professional too. For example depending in who you're emailing like say you're emailing your teacher for help then here's how it'd go:

Dear(teacher name, capitalize, never use first name unless they allow it)

Hello (teacher name), my name is (first and last name) from your (number class) and I was wondering if you could please help me out with (situation, be clear on what you need help with otherwise it won't get through to them)? If you could that would be greatly appreciated!

Sincerely,

(your name first and last)

Answer:

SECTION LEARNING OBJECTIVES

By the end of this section, you will be able to do the following:

Distinguish between static friction and kinetic friction

Solve problems involving inclined planes

Section Key Terms

kinetic friction static friction

Static Friction and Kinetic Friction

Recall from the previous chapter that friction is a force that opposes motion, and is around us all the time. Friction allows us to move, which you have discovered if you have ever tried to walk on ice.

There are different types of friction—kinetic and static. Kinetic friction acts on an object in motion, while static friction acts on an object or system at rest. The maximum static friction is usually greater than the kinetic friction between the objects.

Imagine, for example, trying to slide a heavy crate across a concrete floor. You may push harder and harder on the crate and not move it at all. This means that the static friction responds to what you do—it increases to be equal to and in the opposite direction of your push. But if you finally push hard enough, the crate seems to slip suddenly and starts to move. Once in motion, it is easier to keep it in motion than it was to get it started because the kinetic friction force is less than the static friction force. If you were to add mass to the crate, (for example, by placing a box on top of it) you would need to push even harder to get it started and also to keep it moving. If, on the other hand, you oiled the concrete you would find it easier to get the crate started and keep it going.

Figure 5.33 shows how friction occurs at the interface between two objects. Magnifying these surfaces shows that they are rough on the microscopic level. So when you push to get an object moving (in this case, a crate), you must raise the object until it can skip along with just the tips of the surface hitting, break off the points, or do both. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to move them.

Answer:

The electric ceiling fan was invented in 1882 by engineer and inventor, Philip Diehl. He had earlier invented an electric sewing machine and adapted the motor from this invention to create the ceiling fan. He called his invention the “Diehl Electric Fan” and it was such a success that he soon had many other people competing with him.

Explanation:

Answer:

OPTION A,Larger

HOPE IT HELPS

✅C) Avoid having to use a net ✅

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