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Vlad [161]
4 years ago
3

Between 0 degrees Celsius and 30 degrees Celsius, the volume V (in cubic centimeters) of 1 kg of water at a temperature T is giv

en approximately by the formulaV=999.87−0.06426T+0.0085043T2−0.0000679T3Find the temperature at which water has its maximum density.Temperature = _______________degrees Celsius

Mathematics
1 answer:
Arisa [49]4 years ago
5 0

Answer:

4 degrees Celsius

Step-by-step explanation:

Density = Mass / Volume

Density is Maximum when Volume is Minimum provided that the mass is constant.

As we are given a constant mass of 1 kg, we can use the given formula to tabulate the values of Volume at the values of temperature ranging from 0 to 30 degrees Celsius.

We will then select the value of temperature that provides us with the lowest Volume.

The tabulated results are attached and it can be seen that the minimum Volume (highlighted) is at 4 degrees Celsius.

Hence, we will have the maximum density at 4 degrees Celsius.

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At what rate is a car traveling, if it goes 157.5 miles in 2.5 hours?
Finger [1]

divide total miles by time:

157.5 / 2.5 = 63 miles per hour

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3 years ago
Find the sum of 5/6 +1/8+3/4
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Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the proporti
hodyreva [135]

Answer:

The sample size n = 4225

Step-by-step explanation:

We will use maximum error formula = \frac{z_{a} S.D}{\sqrt{n} }

but we will find sample size "n"

\sqrt{n}  = \frac{z_{a} S.D}{maximum error}

Squaring on both sides , we get

n  = (\frac{z_{a} S.D}{maximum error})^2

Given 99% confidence interval (z value) = 2.56

given maximum error = 0.02

n  = (\frac{z_{a} p(1-p)}{maximum error})^2

n≤ (\frac{2.56X\frac{1}{2} }{0.02}) ^{2}    ( here S.D = p(1-p) ≤ 1/2

on simplification , we get n = 4225

<u>Conclusion</u>:

The sample size of two samples is  n = 4225

<u>verification</u>:-

We will use maximum error formula = \frac{z_{a} S.D}{\sqrt{n} } = \frac{2.56 1/2}{\ \sqrt{4225} }  = 0.0196

substitute all values and simplify we get maximum error is 0.02

4 0
3 years ago
An instructor who taught two sections of Math 161A, the first with 20 students and the second with 30 students, gave a midterm e
uranmaximum [27]

Answer:

<em>The answers are for option (a) 0.2070  (b)0.3798  (c) 0.3938 </em>

Step-by-step explanation:

<em>Given:</em>

<em>Here Section 1 students = 20 </em>

<em> Section 2 students = 30 </em>

<em> Here there are 15 graded exam papers. </em>

<em> (a )Here Pr(10 are from second section) = ²⁰C₅ * ³⁰C₁₀/⁵⁰C₁₅= 0.2070 </em>

<em> (b) Here if x is the number of students copies of section 2 out of 15 exam papers. </em>

<em>  here the distribution is hyper-geometric one, where N = 50, K = 30 ; n = 15 </em>

<em>Then, </em>

<em> Pr( x ≥ 10 ; 15; 30 ; 50) = 0.3798 </em>

<em> (c) Here we have to find that at least 10 are from the same section that means if x ≥ 10 (at least 10 from section B) or x ≤ 5 (at least 10 from section 1) </em>

<em> so, </em>

<em> Pr(at least 10 of these are from the same section) = Pr(x ≤ 5 or x ≥ 10 ; 15 ; 30 ; 50) = Pr(x ≤ 5 ; 15 ; 30 ; 50) + Pr(x ≥ 10 ; 15 ; 30 ; 50) = 0.0140 + 0.3798 = 0.3938 </em>

<em> Note : Here the given distribution is Hyper-geometric distribution </em>

<em> where f(x) = kCₓ)(N-K)C(n-x)/ NCK in that way all these above values can be calculated.</em>

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Answer:

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and so on....

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