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Romashka [77]
2 years ago
15

A watt is a unit of power, which measures energy used per unit time. A kilowatt hour means 1 kilowatt of electricity used contin

uously for 1 hr. The electric utility company in Laura's town charges $0.06001 for each
kilowatt hour used. Laura heats her house with three electric heaters that each use 1200 watts. How much does it cost to heat a house in one day? How many hours what a 75 watt lightbulb have to stay on to result in one dollar for electricity charges?
Mathematics
1 answer:
Bogdan [553]2 years ago
5 0

Answer:

It costs Laura $ 5,184.86 to heat her house one day.

A 75 watt lightbulb should be around 222 hours to consume a dollar in electricity.

Step-by-step explanation:

Given that a kilowatt hour means 1 kilowatt of electricity used continuously for 1 hr, and the electric utility company in Laura's town charges $ 0.06001 for each kilowatt hour used, and Laura heats her house with three electric heaters that each use 1200 watts, to determine how much does it cost to heat a house in one day and how many hours would a 75 watt lightbulb have to stay on to result in one dollar for electricity charges the following calculations must be performed:

(3 x 1200 x 24) x 0.06001 = X

(3600 x 24) x 0.06001 = X

86400 x 0.06001 = X

5,184.86 = X

Therefore, it costs Laura $ 5,184.86 to heat her house one day.

1 watt = 0.001 kilowatt

75 watt = 0.075 kilowatt

1 / 0.075 = 13.333

1 / 0.06001 = 16,663

13,333 x 16,663 = 222.18

Therefore, a 75 watt lightbulb should be around 222 hours to consume a dollar in electricity.

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c = 2d  [not true]

Three less than twice as many cats (as) dogs.

c = 2d - 3

210 combined,

c + d = 210

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The Salk polio vaccine experiment in 1954 focused on the effectiveness of the vaccine in combating paralytic polio. Because it w
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Answer:

Step-by-step explanation:

Hello!

The variables of interest are:

X₁: Number of cases of polio observed in kids that received the placebo vaccine.

n₁= 201299 total children studied

x₁= 110 cases observed

X₂: Number of cases of polio observed in kids that received the experimental vaccine.

n₂= 200745 total children studied

x₂= 33 cases observed

These two variables have a binomial distribution. The parameters of interest, the ones to compare, are the population proportions: p₁ vs p₂

You have to test if the population proportions of children who contracted polio in both groups are different: p₂ ≠ p₁

a)

H₀: p₂ = p₁

H₁: p₂ ≠ p₁

α: 0.05

Z= \frac{(p'_2-p'_1)-(p_2-p_1)}{\sqrt{p'[\frac{1}{n_1} +\frac{1}{n_2} ]} }

Sample proportion placebo p'₁= x₁/n₁= 110/201299= 0.0005

Sample proportion vaccine p'₂= x₂/n₂= 33/200745= 0.0002

Pooled sample proportion p'= (x₁+x₂)/(n₁+n₂)= (110+33)/(201299+200745)= 0.0004

Z_{H_0}= \frac{(0.0002-0.0005)-0}{\sqrt{0.0004[\frac{1}{201299} +\frac{1}{200745} ]} }= -4.76

This test is two-tailed, using the critical value approach, you have to determine two critical values:

Z_{\alpha/2}= Z_{0.025}= -1.96

Z_{1-\alpha /2}= Z_{0.975}= 1.96

Then if Z_{H_0} ≤ -1.96 or if Z_{H_0} ≥ 1.96, the decision is to reject the null hypothesis.

If -1.96 < Z_{H_0} < 1.96, the decision is to not reject the null hypothesis.

⇒ Z_{H_0}= -4.76, the decision is to reject the null hypothesis.

b)

H₀: p₂ = p₁

H₁: p₂ ≠ p₁

α: 0.01

Z= \frac{(p'_2-p'_1)-(p_2-p_1)}{\sqrt{p'[\frac{1}{n_1} +\frac{1}{n_2} ]} }

The value of Z_{H_0}= -4.76 doesn't change, since we are working with the same samples.

The only thing that changes alongside with the level of significance is the rejection region:

Z_{\alpha /2}= Z_{0.005}= -2.576

Z_{1-\alpha /2}= Z_{0.995}= 2.576

Then if Z_{H_0} ≤ -2.576or if Z_{H_0} ≥ 2.576, the decision is to reject the null hypothesis.

If -2.576< Z_{H_0} < 2.576, the decision is to not reject the null hypothesis.

⇒ Z_{H_0}= -4.76, the decision is to reject the null hypothesis.

c)

Remember the level of significance (probability of committing type I error) is the probability of rejecting a true null hypothesis. This means that the smaller this value is, the fewer chances you have of discarding the true null hypothesis. But as you know, you cannot just reduce this value to zero because, the smaller α is, the bigger β (probability of committing type II error) becomes.

Rejecting the null hypothesis using different values of α means that there is a high chance that you reached a correct decision (rejecting a false null hypothesis)

I hope this helps!

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