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irina [24]
3 years ago
10

The price of a 6-minute phone call is 1.80. What is the price of a 12-minute phone call?

Mathematics
2 answers:
Alina [70]3 years ago
7 0

Answer:

21.6 is the price.

Step-by-step explanation:

1.80×12

hoa [83]3 years ago
4 0

Answer:

3.60

Step-by-step explanation:

1.80 x 2

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Use the definition of the derivative to differentiate v=4/2 pie r^3
nata0808 [166]

I suspect 4/2 should actually be 4/3, since 4/2 = 2, while 4/3 would make V the volume of a sphere with radius r. But I'll stick with what's given:

\displaystyle \frac{dV}{dr} = \lim_{h\to0} \frac{2\pi(r+h)^3-2\pi r^3}{h}

\displaystyle \frac{dV}{dr} = 2\pi \lim_{h\to0} \frac{(r^3+3r^2h+3rh^2+h^3)- r^3}{h}

\displaystyle \frac{dV}{dr} = 2\pi \lim_{h\to0} \frac{3r^2h+3rh^2+h^3}{h}

\displaystyle \frac{dV}{dr} = 2\pi \lim_{h\to0} (3r^2+3rh+h^2)

\displaystyle \frac{dV}{dr} = 2\pi \cdot 3r^2 = \boxed{6\pi r^2}

In Mathematica, you can check this result via

D[4/2*Pi*r^3, r]

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3 years ago
I've been looking in the study guide but can't find how to solve it
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The answer is -1/16 or -0.0625, do you need to see work?
6 0
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What is the fraction 18/24 reduce to its lowest terms
zaharov [31]
\frac{18}{24} = \frac{6\times 3}{6\times 4} = \boxed{\frac{3}{4}}
8 0
3 years ago
The average amount of time that students use computers at a university computer center is 36 minutes with a standard deviation o
frosja888 [35]

Answer:

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 36, \sigma = 5

The first step to solve this question is finding the proportion of students which use the computer more than 40 minutes, which is 1 subtracted by the pvalue of Z when X = 40. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{40 - 36}{5}

Z = 0.8

Z = 0.8 has a pvalue of 0.7881.

1 - 0.7881 = 0.2119

So 21.19% of the students use the computer for longer than 40 minutes.

Out of 10000

0.2119*10000 = 2119

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

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3 years ago
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Going to be number two becaus
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