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Kipish [7]
3 years ago
10

How do you solve for m using the equation 5.01cm^3=m/1.2cm^3

Mathematics
1 answer:
4vir4ik [10]3 years ago
5 0
5.01/1.2=m
M=4.175
So your answer for M is 4.175
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1.2x + 3.4=10.6<br> Please help
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1.2x + 3.4 = 10.6
subtract 3.4 from each side
1.2x = 7.2
divide each side by 1.2
x = 6
5 0
3 years ago
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What is the equation in slope-intercept form of a line that is perpendicular to y=13x+2 and passes through the point (3, 1)? Ent
geniusboy [140]

Answer:

y = -1/13x + 1 3/13

Step-by-step explanation:

y = 13x + 2

Perpendicular

y = -1/13x + b

Finding b:

1 = -1/13 * 3 + b

1 = -3/13 + b

1 3/13 = b

y = -1/13x + 1 3/13

Please let me know if I'm wrong :)

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What is the square root of 6 - the square root of 12?<br><br> Please answer fast!
dmitriy555 [2]

Answer:

1.01 or 1.0

Step-by-step explanation:

5 0
3 years ago
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If cheese is $4.40 per kg, what should I pay for 200 g?
Dvinal [7]
1kg = 1000g
1000g/5g= 200g
4.40/5 = 0.88

3 0
3 years ago
At an ocean-side nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water t
grandymaker [24]

Answer:

(a1) The probability that temperature increase will be less than 20°C is 0.667.

(a2) The probability that temperature increase will be between 20°C and 22°C is 0.133.

(b) The probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c) The expected value of the temperature increase is 17.5°C.

Step-by-step explanation:

Let <em>X</em> = temperature increase.

The random variable <em>X</em> follows a continuous Uniform distribution, distributed over the range [10°C, 25°C].

The probability density function of <em>X</em> is:

f(X)=\left \{ {{\frac{1}{25-10}=\frac{1}{15};\ x\in [10, 25]} \atop {0;\ otherwise}} \right.

(a1)

Compute the probability that temperature increase will be less than 20°C as follows:

P(X

Thus, the probability that temperature increase will be less than 20°C is 0.667.

(a2)

Compute the probability that temperature increase will be between 20°C and 22°C as follows:

P(20

Thus, the probability that temperature increase will be between 20°C and 22°C is 0.133.

(b)

Compute the probability that at any point of time the temperature increase is potentially dangerous as follows:

P(X>18)=\int\limits^{25}_{18}{\frac{1}{15}}\, dx\\=\frac{1}{15}\int\limits^{25}_{18}{dx}\,\\=\frac{1}{15}[x]^{25}_{18}=\frac{1}{15}[25-18]=\frac{7}{15}\\=0.467

Thus, the probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c)

Compute the expected value of the uniform random variable <em>X</em> as follows:

E(X)=\frac{1}{2}[10+25]=\frac{35}{2}=17.5

Thus, the expected value of the temperature increase is 17.5°C.

7 0
3 years ago
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