A 50 ohm load dissipates 100 W of power. How much voltage is across the load?

9514 1404 393

  about 70.7 V

Step-by-step explanation:

The relation between power and voltage and resistance in a resistive circuit is ...

  P = V^2/R

so ...

  V = √(PR)

  V = √(100·50) = √5000

  V ≈ 70.7 . . . . volts

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;

idk

step-by-step explanation: