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2 years ago

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A random sample of 25 college statistics textbook prices is obtained and the mean price is computed. To determine the probabilit

y of finding a more extreme mean than the one obtained from this random sample, you would need to refer to:

1 answer:

Lubov Fominskaja [6]2 years ago

4 0

Answer:

We need to refer to the population distribution of all college statistics textbook prices.

Step-by-step explanation:

From the given information;

We are being given the sample size and the mean that has already been computed. Thus, to determine the probability of a more extreme mean, we need the t-test statistics value. In this case, we will need the sample mean, thus we need to refer to the population distribution of all college statistics textbook prices.

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The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature

Karolina [17]

The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

$T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$

Where

$(t) = \ time$

$T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)$

$T_{s} = Surrounding \ temperature$

$T_{0} = Initial \ temperature \ of \ the \ body$

$k = constant$

From the question,

$T_{0} = 86 ^{o}C$

$T_{s} = -20 ^{o}C$

∴ $T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C$

$T_{0} - T_{s} = 106^{o} C$

Therefore, the equation $T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$ becomes

$T_{(t)}=-20+106 e^{kt}$

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time $t = 1 \ hour$ , $T_{(t)} = 58^{o}C$

Then, we can write that

$T_{(1)}=58 = -20+106 e^{k(1)}$

Then, we get

$58 = -20+106 e^{k(1)}$

Now, solve for $k$

First collect like terms

$58 +20 = 106 e^{k}$

$78 =106 e^{k}$

Then,

$e^{k} = \frac{78}{106}$

$e^{k} = 0.735849$

Now, take the natural log of both sides

$ln(e^{k}) =ln( 0.735849)$

$k = -0.30673$

This is the value of the constant $k$

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at $t = 2 \ hours$

From

$T_{(t)}=-20+106 e^{kt}$

Then

$T_{(2)}=-20+106 e^{(-0.30673 \times 2)}$

$T_{(2)}=-20+106 e^{-0.61346}$

$T_{(2)}=-20+106\times 0.5414741237$

$T_{(2)}=-20+57.396257$

$T_{(2)}=37.396257 \ ^{o}C$

$T_{(2)} \approxeq 37.40 \ ^{o}C$

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

Learn more here: brainly.com/question/11689670

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2 years ago

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An object is translated by ( x + 4, y - 2). If one point in the image has the coordinates (5, -3), what would be the coordinates

Nikolay [14]

To find the pre-image we work backwards. Since the coordinates one of the translated image is (5,-3) and this image was translated with the formula (x+4, y-2), then we can just work backwords.
(x-4, y+2) --> (5-4,-3+2) -->(1,-1)
The correct answer is (1,-1)

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3 years ago

Need help again with this

nalin [4]

I think it would be B

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3 years ago

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Please!!!!! help me fill the blanks. I really need it!!! It due tomorrow!!!!

olga nikolaevna [1]

Answer:

The first blank which is in the first row would be 9

The second blank under the 5 is 100

The third blank below the 8 is 160

And the last blank is 240

Step-by-step explanation:

It you do 60 divided by 3, it would be 20, meaning each sweater costs $20.

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2 years ago

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A tragedy with which drug was responsible for the passage of the Federal Food, Drug, and Cosmetic Act of 1938 that required the

matrenka [14]

<span>A. It was the diethylene glycol the drug was put into so it could be given in liquid form that killed people, not the drug</span>

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