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1 year ago

8

Can someone help will give brainlist

1 answer:

Marizza181 [45]1 year ago

5 0

Answer:

720

Step-by-step explanation:

10x9x8

Since they are picking 3 different instruments

please mark me brainliest :)

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Classify the following statement as an example of classical probability, empirical probability, or subjective probability. Exp

AlekseyPX

Answer:

It is a case of classical probability.

Step-by-step explanation:

Since we are selecting a number between 1 and 100 randomly which is divisible by 14 the only favorable cases are 14,28,42,56,70,84,98 which are 7 cases out of 100 total numbers thus the required probability becomes

$P(E)=\frac{7}{100}=0.07$

8 0

2 years ago

Good morning babe I'm sorry for the late reply with quote from the people that have not heard back yet to see you doing anything

WARRIOR [948]

what do you need help with?

8 0

2 years ago

Read 2 more answers

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

6 0

1 year ago

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I don't understand this

Taya2010 [7]

The answer to the problem is B.

7 0

2 years ago

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DO 11. Find the slope of the line that passes through the points<br> (-4, 7/2) and (3, -21/10)

Evgen [1.6K]

Answer:

-4/5

Step-by-step explanation:

5 0

1 year ago