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Karolina [17]
4 years ago
10

A 0.143-Henry Inductor is connected in series with a variable resistor to a 208-volt 400-cycle source. For what value of capacit

ance will the current be (a) 1.04 ampere lagging and (b) 1.04 ampere leading?
Mathematics
1 answer:
aleksandr82 [10.1K]4 years ago
8 0

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

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The function of (x) varies directly with x^2, and f(x)=96 when x=4 what is the value of f(2)
makkiz [27]
\bf \qquad \qquad \textit{direct proportional variation}
\\\\
\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad  y=kx\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad  variation
\end{array}\\\\
-------------------------------

\bf \textit{f(x) varies directly with }x^2\qquad f(x)=kx^2
\\\\\\
\textit{we also know that }
\begin{cases}
f(x)=96\\
x=4
\end{cases}\implies 96=k(4)^2\implies 96=16k
\\\\\\
\cfrac{96}{16}=k\implies 6=k\qquad therefore\qquad \boxed{f(x)=6x^2}
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3 0
3 years ago
Please answer correctly !!!!! Will mark brainliest !!!!!!!!
Vesna [10]

Answer:

k(-7) = = -89

Step-by-step explanation:

k(t) = 10t -19

Let t = -7

k(-7) = 10*-7 -19

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      = -89

7 0
3 years ago
Read 2 more answers
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
4 years ago
Please help, there is a picture.
arlik [135]

Answer:

B (2.5, -0.5)

3 0
2 years ago
Write 4.81 x 10^5 in standard form
tia_tia [17]
481,000
this is the answer because standered form is just asking to write the number out instead of notation
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4 years ago
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