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dangina [55]
3 years ago
5

Find dy/dx for the function y+4x²

Mathematics
1 answer:
docker41 [41]3 years ago
8 0

\dfrac{\partial }{\partial x}(y+4x^2)=0+8x=8x

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A line has a slope of 2/3 and contains the point (-1,-5). Which is an equation of this line?
mario62 [17]

Answer:

y= 2/3x - 13/3

Step-by-step explanation:

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3 0
3 years ago
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sladkih [1.3K]

Answer:

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Step-by-step explanation:

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3 years ago
A ferry will safely accommodate 82 tons of passenger cars. Assume that the meanweight of a passenger car is 2 tons with standard
Shalnov [3]

Answer:

The probability that the maximum safe-weight will be exceeded is <u>0.0455 or 4.55%</u>.

Step-by-step explanation:

Given:

Maximum safe-weight of 37 cars = 82 tons

∴ Maximum safe-weight of 1 car (x) = 82 ÷ 37 = 2.22 tons (Unitary method)

Mean weight of 1 car (μ) = 2 tons

Standard deviation of 37 cars = 0.8 tons

So, standard deviation of 1 car is given as:

\sigma=\frac{0.8}{\sqrt{37}}=0.13

Probability that maximum safe-weight is exceeded, P(x > 2.22) = ?

The sample is normally distributed (Assume)

Now, let us determine the z-score of the mean weight.

The z-score is given as:

z=\frac{x-\mu}{\sigma}\\\\z=\frac{2.22-2}{0.13}\\\\z=\frac{0.22}{0.13}=1.69

Now, finding P(x > 2.22) is same as finding P(z > 1.69).

From the z-score table of normal distribution curve, the value of area under the curve for z < 1.69 is 0.9545.

But we need the area under the curve for z > 1.69.

So, we subtract from the total area. Total area is 1 or 100%.

So, P(z > 1.69) = 1 - P(z < 1.69)

P(z>1.69)=1-0.9545=0.0455\ or\ 4.55\%

Therefore, the probability that the maximum safe-weight will be exceeded is 0.0455 or 4.55%.

8 0
4 years ago
Plssss helppp me plsss
adelina 88 [10]
6 I think but maybe 1
4 0
3 years ago
Read 2 more answers
12. Find the values of x and y in the diagram below.
ehidna [41]

Answer:

x = 15

y = 19

8 0
3 years ago
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