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1 year ago

Q19. When a resistor is placed across 230M supply (de) the

Mathematics

1 answer:

Vera_Pavlovna [14]1 year ago

6 0

The value of the resistor that must be connected in parallel is 57.47 ohms

<h3>How to determine the resistor connected across the 230 V supply</h3>

Voltage (V) = 230 V
Current (I) = 12 A
Resistor 1 (R₁) =?

V = IR

230 = 12 × R₁

Divide both sides by 12

R₁ = 230 / 12

R₁ = 19.17 ohms

<h3>How to determine the total resistance</h3>

Voltage (V) = 230 V
Current (I) = 16 A
Total resistance (Rₜ) =?

V = IR

230 = 16 × Rₜ

Divide both sides by 12

Rₜ = 230 / 16

Rₜ = 14.375 ohms

<h3>How to determine the resistor connected in paralle</h3>

Resistor 1 (R₁) = 19.17 ohms
Total resistance (Rₜ) = 14.375 ohms
Resistor 2 (R₂) = ?

1/Rₜ = 1/R₁ + 1/R₂

Collect like terms

1/R₂ = 1/Rₜ - 1/R₁

R₂ = (Rₜ × R₁) / (R₁ - Rₜ)

R₂ = (14.375 × 19.17) / (19.17 - 14.375)

R₂ = 57.47 ohms

Learn more about Ohm's law:

brainly.com/question/796939

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1 year ago

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2 years ago

Based on the Nielsen ratings, the local CBS affiliate claims its 11 p.m. newscast reaches 41% of the viewing audience in the are

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Answer:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

$z=\frac{0.36 -0.41}{\sqrt{\frac{0.41(1-0.41)}{100}}}=-1.017$

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

$\hat p=0.36$ estimated proportion with the survey

$p_o=0.41$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is lower than 0.41.:

Null hypothesis: $p\geq 0.41$

Alternative hypothesis: $p < 0.41$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

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Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.36 -0.41}{\sqrt{\frac{0.41(1-0.41)}{100}}}=-1.017$

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