Question 2 Unsaved Simplify. 3c812c12

Answers

Answer 1

Answer: I think the answer would be that the dimplified form of 3c8/12c12 is
c^20
———
4

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From a random sample of 41 teens, it is found that on average they spend 43.1 hours each week online with a population standard deviation of 5.91 hours. What is the 90% confidence interval for the amount of time they spend online each week?

Answers

Answer:

Step-by-step explanation:

We want to determine a 90% confidence interval for the mean amount of time that teens spend online each week.

Number of sample, n = 41

Mean, u = 43.1 hours

Standard deviation, s = 5.91 hours

For a confidence level of 90%, the corresponding z value is 1.645. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean +/- z ×standard deviation/√n

It becomes

43.1 ± 1.645 × 5.91/√41

= 43.1 ± 1.645 × 0.923

= 43.1 ± 1.52

The lower end of the confidence interval is 43.1 - 1.52 =41.58

The upper end of the confidence interval is 43.1 + 1.52 =44.62

Therefore, with 90% confidence interval, the mean amount of time that teens spend online each week is between 41.58 and 44.62

6-2d=42

What is D please help me find what it is

Answers

Answer: -18

Step-by-step explanation:

if 6-2d=42

just subtract 6 from both sides

-2d=42-6

-2d=36 now divide by 2

-d=18

multiply by -1

d=-18

Answer: d=-18

Step-by-step explanation:

6-2d=42

First subtract 6 from both sides

-2d=36

Now divide both sides by -2 to get d alone

d=-18

A patient is given a 50 mg dose of medicine the medicines affective Ness decreases every hour at a constant rate of 40% what is the exponential decay function that models this scenario how much medicine will be left in the patient's system after two hours

Answers

The following formula is used to find the answer.
D = 50 mg (0.6^n)
D is the dosage
n is at any hour
Using this formula and solving the equation for it, the answer is 18.

A 0.143-Henry Inductor is connected in series with a variable resistor to a 208-volt 400-cycle source. For what value of capacitance will the current be (a) 1.04 ampere lagging and (b) 1.04 ampere leading?

Answers

Answer:

A.)359.2, B.)2.5 uf

Step-by-step explanation:

E / I = R

208 / 1.04 = 200 ohms

2*pi*f*L = Xl

6.28*400*.143 = 359.2 ohm

1 / (2*pi*f*Xc) = c

1 /(6.28*400*159.2) = 2.5 uf

Final answer:

The question asked for the value of capacitance that causes the current in an AC circuit to lag or lead. This situation occurs at resonance when the reactance of the inductor equals that of the capacitor. The calculation of capacitance utilizes the resonance formula, and both given scenarios (a and b) were calculated using provided circuit properties.

Explanation:

The subject of this question involves the principles of alternating current (AC) circuits which includes concepts of inductance, capacitance, and impedance. Particularly, the question is asking to find the value of the capacitor (capacitance) that will result in a current that is (a) lagging or (b) leading in an AC circuit with a given inductor connected in series with a resistor and a power source.

Resonance in AC circuits

When the reactance of the inductor, L, equals the reactance of the capacitor, C, the circuit attains a state called resonance. At resonance, the total impedance of the circuit is at its minimum, hence, the current is at its maximum. This happens when the current leads or lags the voltage.