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miskamm [114]
3 years ago
15

C=5/9 (F-32). Find C for F=5°

Mathematics
1 answer:
olasank [31]3 years ago
6 0

Answer:

-15

Step-by-step explanation:

\frac{5}{9} ((5)-32)\\\frac{5}{9}(-27)\\-15

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50 x 1% = 0.5
because 1% is equal to 1/100 so 50 x 1/100 is equal to 50/100 or 1/2 which is equal to 0.5
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What is the diameter of the following circle (x+4)^2 + (y-9)^2 = 18
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3 0
3 years ago
A random sample of size n1= 25, taken from a normal population with a standard deviation σ1= 5, has a mean X1= 80. A second rand
sesenic [268]

Answer:

The 94% confidence interval would be given by 2.898 \leq \mu_1 -\mu_2 \leq 7.102

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X_1 =80 represent the sample mean 1

\bar X_2 =75 represent the sample mean 2

n1=25 represent the sample 1 size  

n2=36 represent the sample 2 size  

\sigma_1 =5 sample standard deviation for sample 1

\sigma_2 =3 sample standard deviation for sample 2

\mu_1 -\mu_2 parameter of interest.

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}} (1)  

The point of estimate for \mu_1 -\mu_2 is just given by:

\bar X_1 -\bar X_2 =80-75=5

Since the Confidence is 0.94 or 94%, the value of \alpha=0.06 and \alpha/2 =0.03, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.03,0,1)".And we see that z_{\alpha/2}=1.88  

The standard error is given by the following formula:

SE=\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}

And replacing we have:

SE=\sqrt{\frac{5^2}{25}+\frac{3^2}{36}}=1.118

Confidence interval

Now we have everything in order to replace into formula (1):  

5-1.88\sqrt{\frac{5^2}{25}+\frac{3^2}{36}}=2.898  

5+1.8\sqrt{\frac{6^2}{36}+\frac{7^}{49}}=7.102  

So on this case the 94% confidence interval would be given by 2.898 \leq \mu_1 -\mu_2 \leq 7.102  

7 0
3 years ago
Many utility companies promote energy conservation by offering discount rates to consumers who keep their energy usage below cer
aleksley [76]

Answer:

a. 0.1681 = 16.81% probability that all five qualify for the favorable rate.

b. 0.5283 = 52.83% probability that at least four qualify for the favorable rates

Step-by-step explanation:

For each Puerto Rico resident, there are only two possible outcomes. Either they qualify for discounted rates, or they do not. The probability of a person in the sample qualifying for discounted rates is independent of any other person in the sample. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

70% of the island residents of Puerto Rico have reduced their electricity usage sufficiently to qualify for discounted rates.

This means that p = 0.7

Five residential subscribers are randomly selected from San Juan, Puerto Rico

This means that n = 5

a. All five qualify for the favorable rate

This is P(X = 5). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 5) = C_{5,5}.(0.7)^{5}.(0.3)^{0} = 0.1681

0.1681 = 16.81% probability that all five qualify for the favorable rate.

b. At least four qualify for the favorable rates

This is

P(X \geq 4) = P(X = 4) + P(X = 5)

So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 4) = C_{5,4}.(0.7)^{4}.(0.3)^{1} = 0.3602

P(X = 5) = C_{5,5}.(0.7)^{5}.(0.3)^{0} = 0.1681

P(X \geq 4) = P(X = 4) + P(X = 5) = 0.3602 + 0.1681 = 0.5283

0.5283 = 52.83% probability that at least four qualify for the favorable rates

5 0
3 years ago
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