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3 years ago

PV=nRT, solve for T

Mathematics

2 answers:

EastWind [94]3 years ago

8 0

$\huge\underline{\underline{\mathfrak{\red{A}\green{n}\pink{s}\orange{w}\blue{e}{r}}}}$

$$P \cdot V = n \cdot R \cdot T$$

Dividing both sides by n · R:

$$\displaystyle \frac{P \cdot V}{n \cdot R} =T$$

$$\rule[225]{225}{2}$$

Katen [24]3 years ago

5 0

Answer:

$\huge \boxed{T= \frac{P V}{n R} }$

$\rule[225]{225}{2}$

Step-by-step explanation:

$P \cdot V = n \cdot R \cdot T$

Dividing both sides by n · R:

$\displaystyle \frac{P \cdot V}{n \cdot R} =T$

$\rule[225]{225}{2}$

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What is the slope of the line represented y= 5x -7

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When using the rational root theorem, which of the following is a possible root of the polynomial function below f(x)=x^3-5x^2-1

uranmaximum [27]

Answer:

$\Large \boxed{\sf \ \ 7 \ \ }$

Step-by-step explanation:

Hello, please consider the following.

The polynomial function is

$x^3-5x^2-12x+14$

The rational root theorem states that each rational solution

$x=\dfrac{p}{q}$

, written in irreducible fraction, satisfies the two following:

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q is a factor of the leading coefficient

In this example, the constant term is 14 and the leading coefficient is 1. It means that p is a factor of 14 and q a factor of 1.

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14 = 2 * 7

Finally, the possible rational roots of this expression are :

and we need to test for negative ones too

-1

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4 years ago

In a study of the impact of smoking on birth weight, researchers analyze birth weights (in grams) for babies born to 189 women w

kompoz [17]

Answer:

c. Smoking is associated with lower birth weights. When smokers are compared to nonsmokers, we are 95% confident that the mean weight of babies born to nonsmokers will be between 76.5 grams to 486.9 grams more than the mean birth weight of babies born to smokers.

Step-by-step explanation:

The difference in mean birth weights (nonsmokers minus smokers) is 281.7 grams with a margin of error of 205.2 grams with 95% confidence.

Then, we know that the 95% confidence interval is (76.5, 486.9)

$LL=M-MOE=281.7-205.2=76.5\\\\UL=M-MOE=281.7+205.2=486.9$

a. We are 95% confident that smoking causes lower birth weights by an average of between 76.5 grams to 486.9 grams.

False, it interprets the confidence interval as a probability for individual cases. The confidence interval is a range for the population mean.

b. There is a 95% chance that if a woman smokes during pregnancy her baby will weigh between 76.5 grams to 486.9 grams less than if she did not smoke.

False, it interprets the confidence interval as a probability for individual cases. The confidence interval is a range for the population mean.

c. Smoking is associated with lower birth weights. When smokers are compared to nonsmokers, we are 95% confident that the mean weight of babies born to nonsmokers will be between 76.5 grams to 486.9 grams more than the mean birth weight of babies born to smokers.

True.

d. With such a large margin of error, this study does not suggest that there is a difference in mean birth weights when we compare smokers to nonsmokers.

There is enough evidence, as the lower bound of the confidence is positive. This means that there is only a probability of 0.05/2=0.025 that the true mean weight difference is smaller than 76.5 grams.

6 0

4 years ago

Y is a random variable that is distributed N(-16, 1.21). Find k such that Prob(-15.043 < Y ≤ k) = 0.1546. (Round your answer

IgorLugansk [536]

Transform Y to Z, which is distributed N(0, 1), using the formula

Y = µ + σZ

where µ = -16 and σ = 1.21.

Pr[-15.043 < Y ≤ k] = 0.1546

Pr[(-15.043 + 16)/1.21 < (Y + 16)/1.21 ≤ (k + 16)/1.21] = 0.1546

Pr[0.791 < Z ≤ (k + 16)/1.21] ≈ 0.1546

Pr[Z ≤ (k + 16)/1.21] - Pr[Z < 0.791] = 0.1546

Pr[Z ≤ (k + 16)/1.21] = 0.1546 + Pr[Z < 0.791]

Pr[Z ≤ (k + 16)/1.21] ≈ 0.1546 + 0.786

Pr[Z ≤ (k + 16)/1.21] ≈ 0.940

Take the inverse CDF of both sides (Φ(x) denotes the CDF itself):

(k + 16)/1.21 ≈ Φ⁻¹ (0.940) ≈ 1.556

Solve for k :

k + 16 = 1.21 • 1.556

k ≈ -14.118

8 0

3 years ago

PV=nRT, solve for T​

PV=nRT, solve for T