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3 years ago

Which of the following elements of the CIA triad refers to maintaining and assuring the accuracy of data over its life-cycle?

Engineering

1 answer:

kenny6666 [7]3 years ago

4 0

Answer:

Integrity: involves maintaining and assuring the accuracy of data over its life-cycle

Explanation:

Confidentiality: This is a CIA triad designed to prevent sensitive information from reaching the wrong people, while making sure that the right people have access to it.

Integrity: This is a CIA triad that involves maintaining the consistency, accuracy, and trustworthiness of data over its entire life cycle.

Availability: This is a CIA triad that involves hardware repairs and maintaining a correctly functioning operating system environment that is free of software conflicts.

Authentication:This is a security control that is used to protect the system with regard to the CIA properties.

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The Environmental Protection Agency (EPA) has standards and regulations that says that the lead level in soil cannot exceed the

DENIUS [597]

Answer:

See below

Explanation:

Check One-Sample T-Interval Conditions

Random Sample? √

Sample Size ≥30? √

Independent? √

Population Standard Deviation Unknown? √

One-Sample T-Interval Information

Formula --> $CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})$
Sample Mean --> $\bar{x}=390.25$
Critical Value --> $t^*=2.0096$ (given $df=n-1=50-1=49$ degrees of freedom at a 95% confidence level)
Sample Size --> $n=50$
Sample Standard Deviation --> $S_x=30.5$

Problem 1

The critical t-value, as mentioned previously, would be $t^*=2.0096$ , making the 95% confidence interval equal to $CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})=390.25\pm2.0096(\frac{30.5}{\sqrt{50}})\approx\{381.5819,398.9181\}$

This interval suggests that we are 95% confident that the true mean levels of lead in soil are between 381.5819 and 398.9181 parts per million (ppm), which satisfies the EPA's regulated maximum of 400 ppm.

3 0

2 years ago

A lake with constant volume 1.1 x 10^6 m^3 is fed by a stream with a non-conservative pollutant of 2.3 mg/L and flow rate 35 m^3

Jet001 [13]

Answer:

12.84 mg/L

Explanation:

We are given;

Volume of lake; V = 1.1 x 10^(6) m³

decay coefficient; K = 0.10/day = 0.1/(24 × 60 × 60) /s = 0.00000115741 /s

Factory rate: Q_f = 4.3 m³/s

Factory concentration: C_f = 100 mg/L

Stream rate: Q_s = 34 m³/s

Stream Concentration: C_s = 2.3 mg/L

Now, to find the steady state concentration of pollutant in the lake, we will use the formula;

(Q_s•C_s) + (Q_f•C_f) = (Q_f + Q_s)C_L + (KV•C_L)

Where C_L is the steady state concentration of pollutant in the lake.

Thus, making C_L the subject, we have;

C_L = [(Q_s•C_s) + (Q_f•C_f)]/(Q_f + Q_s + K•V)

Plugging in the relevant values gives;

C_L = ((34 × 2.3) + (4.3 × 100))/(4.3 + 34 + (0.00000115741 × 1.1 × 10^(6)))

C_L = 12.84 mg/L

4 0

3 years ago

A series AC circuit contains a resistor, an inductor of 250 mH, a capacitor of 4.40 µF, and a source with ΔVmax = 240 V operatin

slega [8]

Answer:

Explanation:

Inductance = 250 mH = 250 / 1000 = 0.25 H

capacitance = 4.40 µF = 4.4 × 10⁻⁶ F ( µ = 10⁻⁶)

ΔVmax = 240, f frequency = 50Hz and I max = 110 mA = 110 /1000 = 0.11A

a) inductive reactance = 2πfl = 2 × 3.142 × 50 × 0.25 H =78.55 ohms

b) capacitive reactance = $\frac{1}{2\pi fC}$ = 1 / ( 2 × 3.142× 50 × 4.4 × 10⁻⁶ ) = 723.34 ohms

c) impedance = $\frac{Vmax}{Imax}$ = 240 / 0.11 = 2181.82 ohms

7 0

3 years ago

Read 2 more answers

In one study the critical stress intensity factor for human bone was calculated to be 4.05 MN/m3/2. If the value of Y in Eq. (2.

Diano4ka-milaya [45]

Where is Eq.(28) ?? You should show it to find the result

6 0

3 years ago

A cylindrical insulation for a steam pipe has an inside radius rt = 6 cm, outside radius r0 = 8 cm, and a thermal conductivity k

goldfiish [28.3K]

Answer:

heat loss per 1-m length of this insulation is 4368.145 W

Explanation:

given data

inside radius r1 = 6 cm

outside radius r2 = 8 cm

thermal conductivity k = 0.5 W/m°C

inside temperature t1 = 430°C

outside temperature t2 = 30°C

to find out

Determine the heat loss per 1-m length of this insulation

solution

we know thermal resistance formula for cylinder that is express as

Rth = $\frac{ln\frac{r2}{r1}}{2 \pi *k * L}$ .................1

here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity

heat loss is change in temperature divide thermal resistance

Q = $\frac{t1- t2}{\frac{ln\frac{r2}{r1}}{2 \pi *k * L}}$

Q = $\frac{(430-30)*(2 \pi * 0.5 * 1}{ln\frac{8}{6} }$

Q = 4368.145 W

so heat loss per 1-m length of this insulation is 4368.145 W

4 0

3 years ago