The picture shows the $481 four friends worked together to earn.

Determine the currents in the picture

Elenna [48]

Answer:

The value of currents are $I_1=0A$ , $I_2=1A$ and $I_3=1A$ .

Step-by-step explanation:

The Ohm's law states that

$V=IR$

it is given that the V₁=3V and V₂=4V.

In a parallel circuit:

Voltage is same in each component

Sum of the currents equals to the total current that flows.

$I_2=I_1+I_3$ .... (1)

Equation for first loop is,

$V_1=4I_1+3I_2$

$3=4I_1+3I_2$ ..... (2)

Equation for second loop is,

$V_2=1I_3+3I_2$

$4=1I_3+3I_2$ ..... (3)

On solving (1), (2) and (3) we get

$I_1=0$

$I_2=1$

$I_3=1$

Therefore the value of currents are $I_1=0A$ , $I_2=1A$ and $I_3=1A$ .

6 0

3 years ago

Solve the problem. Use the Central Limit Theorem.The annual precipitation amounts in a certain mountain range are normally distr

bazaltina [42]

Answer:

0.8944 = 89.44% probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches.

Step-by-step explanation:

To solve this question, we use the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .