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3 years ago

2. Solve the following equations for 0° Sx < 180°.

Mathematics

1 answer:

diamong [38]3 years ago

5 0

Answer:

See below in bold.

Step-by-step explanation:

(1) cosec ( x + 10) = 3

Now cosec x = 1/sin x so

1 / sin (x + 10) = 3

3 sin (x + 10) = 1

sin (x + 10) = 1/3

x + 10 = 19.47 , 160.53 degrees

x = 9.47, 150.53 degrees.

(ii) cot (x - 30) =0.45

cot (x - 30)= 1 /tan (x- 30) so we have

tan (x - 30) = 1 / 0.45 = 2.2222

x - 30 = 65.77 degrees

x = 95.77 degrees.

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If the null hypothesis of an experiment is “The true mean weight of the piglets is at least 39lbs” what is the alternate hypothe

AlexFokin [52]

Answer:

“The true mean weight of the piglets is at most 39lbs”would be the alternate hypothesis.

Step-by-step explanation:

Given that the null hypothesis of an experiment is “The true mean weight of the piglets is at least 39lbs”

We have to find the alternate hypothesis.

In hypothesis testing, alternate is the opposite of null hypothesis. If null hypothesis supports one statement alternate hypothesis opposes it. In other words, alternate hypothesis would be the negative of the claim of null hypothesis.

Here the null hypothesis of an experiment is “The true mean weight of the piglets is at least 39lbs”

i.e H0: $x bar\geq 39 lbs$

So alternate hypothesis

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3 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
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Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

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3 years ago

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icang [17]

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(Hope this helps)

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3 years ago

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Paha777 [63]

Answer:

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tester [92]

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