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ICE Princess25 [194]
3 years ago
7

The equation d=13/v shows that the density of a particular substance equals a mass of 13 grams divided by the volume of the subs

tance. What happens to the density as the volume approaches 0?
Mathematics
1 answer:
stich3 [128]3 years ago
3 0
D = M/V where M=mass and V = volume

Assume the Mass is 1 kg and the Volume = 1 cm³, so d= 1/1 kg/cm³

for V=0.1 → d = 1/0.1 = 10 kg.cm³
for V = 0.01 → d = 100.kg.cm³
for V = 0.001 → d = 1000.kg.cm³
for V = 0.000001 → d = 1,000,000.kg.cm³
and when Volume → 0. density →∞
.
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Step-by-step explanation:

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2 years ago
Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean. See Attached Excel for Da
alexgriva [62]

Answer:

The 99% confidence interval for the mean germination time is (12.3, 19.3).

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>Recorded here are the germination times (in days) for ten randomly  chosen seeds of a new type of bean: 18, 12, 20, 17, 14, 15, 13, 11, 21, 17. Assume that the population germination time is normally distributed. Find the 99% confidence interval for the mean germination time.</em>

<em />

We start calculating the sample mean M and standard deviation s:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{10}(18+12+20+17+14+15+13+11+21+17)\\\\\\M=\dfrac{158}{10}\\\\\\M=15.8\\\\\\

s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{9}((18-15.8)^2+(12-15.8)^2+(20-15.8)^2+. . . +(17-15.8)^2)}\\\\\\s=\sqrt{\dfrac{101.6}{9}}\\\\\\s=\sqrt{11.3}=3.4\\\\\\

We have to calculate a 99% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=15.8.

The sample size is N=10.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{3.4}{\sqrt{10}}=\dfrac{3.4}{3.162}=1.075

The degrees of freedom for this sample size are:

df=n-1=10-1=9

The t-value for a 99% confidence interval and 9 degrees of freedom is t=3.25.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=3.25 \cdot 1.075=3.49

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 15.8-3.49=12.3\\\\UL=M+t \cdot s_M = 15.8+3.49=19.3

The 99% confidence interval for the mean germination time is (12.3, 19.3).

8 0
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