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MrRissso [65]
3 years ago
7

The travel agent says the average high temperature at the resort is 25 degrees celsius. what is the equivalent fahrenheit temper

ature? solve the problem by substituting the appropriate value(s) of the known variable(s) in the formula you found in the previous step. round your answer to two decimal places, if necessary.
Mathematics
1 answer:
Anon25 [30]3 years ago
5 0

Answer:

The equivalent is 77°F

Step-by-step explanation:

Given

Temperature= 25 °C

Required

Convert to Fahrenheit (°F )

To convert from Celsius to Fahrenheit, we simply make use of the following formula:

\°F = (\°C * 9/5) + 32

Substitute 25 for °C

\°F = (25* 9/5) + 32

\°F = (225/5) + 32

Open bracket

\°F = 45 + 32

\°F = 77

Hence;

<em>The equivalent is 77°F </em>

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Answer:

Below

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x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

Step-by-step explanation:

5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}

Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}

\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x

\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

3 0
3 years ago
Weatherwise magazine is published in association with the American Meteorological Society. Volume 46, Number 6 has a rating syst
lisabon 2012 [21]

Answer:

a)  c) μ = 16.4.

b)  d) μ > 16.4.

c) a) μ < 16.4.

d) c) μ ≠ 16.4.

e)  d) right; left; both.

Step-by-step explanation:

Question a:

Test if it is getting worse, so at the alternative hypothesis we test if the mean is of greater than 16.4 inches, but at the null hypothesis we test if it is still of 16.4 options, so option C.

Question b:

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Question c:

Dying down, so if the mean is lower than 16.4 inches, so option a.

Question d:

Don't know, so just test if it is different, which includes both lower or greater, so the correct answer is given by option c.

Question e:

Test if more -> right, so on question b) is a right tailed test.

Test if less -> left, so on question c) is a left tailed test.

Different -> both sides, so on question d) it is a two-tailed test.

Thus the correct answer is given by option d.

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