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Sladkaya [172]
3 years ago
10

A lake near the Arctic Circle is covered by a 222-meter-thick sheet of ice during the cold winter months. When spring arrives, t

he warm air gradually melts the ice, causing its thickness to decrease at a constant rate. After 333 weeks, the sheet is only 1.251.251, point, 25 meters thick. Let S(t)S(t)S, left parenthesis, t, right parenthesis denote the ice sheet's thickness SSS (measured in meters) as a function of time ttt (measured in weeks).
Mathematics
2 answers:
Ivanshal [37]3 years ago
7 0

S(t) = 3.82 - 0.2t


Step-by-step explanation:


After 7 weeks, the ice is 2.42 meters thick. The ice loses 0.2 meters of thickness per week; this means on the 7th week, it has lost 7(0.2) = 1.4 meters of thickness.


This means the ice started at 2.42+1.4 = 3.82 meters thick.


Our function will start at the original thickness of the ice, 3.82 meters.


Since the ice is losing thickness, we will subtract; it loses at a rate of 0.2 meters per week (t), which gives us 0.2t. This is subtracted from the original, 3.82 meters, giving us


S(t) = 3.82 - 0.2t



Delvig [45]3 years ago
3 0

Answer:

S(t)=−0.25t+2

Step-by-step explanation:

if you're on khan academy this is correct!

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Katen [24]

Answer

2 1/8

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2 years ago
The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a st
Alinara [238K]

Answer:

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = \frac{X - \mu}{\sigma}

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer, so \mu = 0.5, \sigma = 0.05.

What is the probability that a line width is greater than 0.62 micrometer?

That is P(X > 0.62)

So

Z = \frac{X - \mu}{\sigma}

Z = \frac{0.62 - 0.5}{0.05}

Z = 2.4

Z = 2.4 has a pvalue of 0.99180.

This means that P(X \leq 0.62) = 0.99180.

We also have that

P(X \leq 0.62) + P(X > 0.62) = 1

P(X > 0.62) = 1 - 0.99180 = 0.0082

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

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Answer:

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Step-by-step explanation:

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X³+y³+z³=<br><br> A. 42?<br> B. 45?
kramer

Answer:

42

Step-by-step explanation:

5 0
3 years ago
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