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2 years ago

Will mark brainliest for whoever answers correctly

Mathematics

1 answer:

siniylev [52]2 years ago

6 0

I think it is D:{5,-infinity}

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The temperature, T, of a given mass of gas varies inversely with its Volume,V. The temperature of 98 cm ³ of a certain gas is 28

Naily [24]

Answer:

T₂ = 421.4 K

Step-by-step explanation:

The temperature, T, of a given mass of gas varies inversely with its Volume,V.

$T\propto \dfrac{1}{V}\\\\V_1T_1=V_2T_2$

We have,

V₁ = 98 cm³, T₁ = 28° C = (28+273) K = 301 K

V₂ = 70 cm₃

We need to find the new temperature of the gas. Using the above relation to find T₂.

$T_2=\dfrac{V_1T_1}{V_2}\\\\T_2=\dfrac{98\times 301}{70}\\\\T_2=421.4\ K$

Hence, the required temperature of the gas is 421.4 K.

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1 year ago

Simplify the expression given below.

Nikitich [7]

I don’t see a expression...

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2 years ago

The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean

LiRa [457]

Answer:

$P(X\geq 3.4)=0.0228$

Step-by-step explanation:

Given the mean is 3.2, standard deviation is 0.8 and the sample size is 64.

-We calculate the probability of a mean of 3.4 as follows:

#First determine the z-value:

$z=\frac{\bar X -\mu}{\sigma/\sqrt{n}}\\\\=\frac{3.4-3.2}{0.8/\sqrt{64}}\\\\=2.000$

#We then determine the corresponding probability on the z tables:

$Z(X\geq 3.4)=1-P(X$

Hence, the probability of obtaining a sample mean this large or larger is 0.0228

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2 years ago

Equations with parentheses may require the use of the distributive property to be solve

Mamont248 [21]

If this is a T/F question, then your answer is True .-.

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2 years ago

Can someone assist me with this, I'm having problems with this

nirvana33 [79]

It’s true!! Hope this helped!!!!

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2 years ago