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3 years ago

7

A chemical compound has a freezing point of -5°C and a boiling point of 97°C. What is the difference between the freezing point

and the boiling point of the chemical compound? °C

1 answer:

rusak2 [61]3 years ago

7 0

Answer:

102°C

Step-by-step explanation:

The chemical compound has a freezing point of -5°C and a boiling point of 97°C.

Since the freezing point is the lower temperature and the boiling point is the higher temperature, the difference between the two temperatures will be subtraction of the freezing point from the boiling point. That is:

97 - (-5) = 97 + 5 = 102°C

The difference between the freezing point and the boiling point is 102°C

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Kite WXYZ is graphed on a coordinate plane. What is the approximate perimeter of the kite? Round to the nearest tenth. 10.6 unit

Archy [21]

<u>Answer-</u>

<em>Perimeter of the kite is </em><em>16.2 units</em>

<u>Solution-</u>

As WXYZ is a kite, so two disjoint pairs of consecutive sides are congruent, i.e WX=XY and WZ=ZY

So, perimeter of the kite WXYZ is,

$=2(\overline{WX}+\overline{WZ})$

And

$\overline{WX}=\sqrt{(1-3)^2+(1-4)^2}=\sqrt{(-2)^2+(-3)^2}=\sqrt{4+9}=\sqrt{13}$

$\overline{WZ}=\sqrt{(1-3)^2+(1+3)^2}=\sqrt{(-2)^2+(4)^2}=\sqrt{4+16}=\sqrt{20}$

So, perimeter will be,

$P=2(\sqrt{13}+\sqrt{20})=16.15\approx 16.2\ units$

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2 years ago

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Please help me, i promise its worth it!!!

Rashid [163]

$\\ \sf\longmapsto 2(L+B)=55$

$\\ \sf\longmapsto 2(\dfrac{4}{3}x+x)=55$

$\\ \sf\longmapsto \dfrac{8}{3}x+2x=55$

$\\ \sf\longmapsto \dfrac{8x+6x}{3}=55$

$\\ \sf\longmapsto \dfrac{14x}{3}=55$

$\\ \sf\longmapsto 14x=165$

$\\ \sf\longmapsto x=11.78$

$\\ \sf\longmapsto x\approx 12$

Now

B=x=12
L=4/3(12)=4(4)=16

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2 years ago

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The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

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3 years ago

A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. Assume that the distr

andrew11 [14]

Answer:hiiiiiiiii

Step-by-step explanation:

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3 years ago

A manufacturing company produces parts that are good (90%), partially defective (2%), or completely defective (8%). These parts

Schach [20]

Answer:

The probability that a part is good given that it passed the inspection machine is P=0.978.

Step-by-step explanation:

As the inspection machine detect and discard any part that is completely defective, only the good and partially defective parts passed this inspection.

Then, if we have:

Probability of being a good part P(G)=0.90

Probability of being a patially defective part P(P)=0.02

Probability of being a completely defective part P(D)=0.08.

Probability of passing the inspection machine = 1-P(D)=1-0.08=0.92

Then, the probability of having a good part, given that it passed the inspection machine is:

$P(G|Pass)=P(G)/P(Pass)=0.90/0.92=0.978$

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3 years ago