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3 years ago

Differentiate the following function. y=x(x^2+5)^2

1 answer:

8 0

First, expand to make it easier
y=x(x^2+5)(x^2+5)
y=x(x^4+10x^2+25)
y=x^5+10x^3+25x
differentiation
y'=5x^4+30x^2+25 is the derivitive

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3 years ago

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Y = -x + 2 y = 3x- 4 Whats x and y

miskamm [114]

Answer:

x=2 y=2

Step-by-step explanation:

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3 years ago

The SAT scores have an average of 1200 with a standard deviation of 60. A sample of 36 scores is selected. a) What is the probab

LiRa [457]

Answer:

a) 0.0082

b) 0.9987

c) 0.9192

d) 0.5000

e) 1

Step-by-step explanation:

The question is concerned with the mean of a sample.

From the central limit theorem we have the formula:

$z=\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$

a) $z=\frac{1224-1200}{\frac{60}{\sqrt{36} } }=2.40$

The area to the left of z=2.40 is 0.9918

The area to the right of z=2.40 is 1-0.9918=0.0082

$\therefore P(\bar X\:>\:1224)=0.0082$

b) $z=\frac{1230-1200}{\frac{60}{\sqrt{36} } }=3.00$

The area to the left of z=3.00 is 0.9987

$\therefore P(\bar X\:$

c) The z-value of 1200 is 0

The area to the left of 0 is 0.5

$z=\frac{1214-1200}{\frac{60}{\sqrt{36} } }=1.40$

The area to the left of z=1.40 is 0.9192

The probability that the sample mean is between 1200 and 1214 is

$P(1200\:$

d) From c) the probability that the sample mean will be greater than 1200 is 1-0.5000=0.5000

e) $z=\frac{73.46-1200}{\frac{60}{\sqrt{36} } }=-112.65$

The area to the left of z=-112.65 is 0.

The area to the right of z=-112.65 is 1-0=1

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3 years ago

Two pools are being drained. To start, the first pool had 3850 liters of water and the second pool had 4370 liters of water. Wat

SashulF [63]

Answer:

(a)

Amount of water in the first pool (in liters) = 3850 liters - 33 liters per minute * x minutes

Amount of water in the second pool (in liters) = 4370 liters - 43 liters per minute * x minutes

(b)

3850 liters - 33 liters per minute * x minutes= 4370 liters - 43 liters per minute * x minutes

Step-by-step explanation:

(a) With x being the minutes after which you want to calculate the amount of water in the pool, to calculate this amount of water you must subtract the water drained from the initial amount of water that the pool contains.

Knowing that, for example, the water from the first pool drains at a rate of 33 liters per minute, then after x minutes, the total water drained will be 33 liters per minute * x minutes. Then:

Amount of water in the first pool (in liters) = 3850 liters - 33 liters per minute * x minutes

Reasoning in the same way:

Amount of water in the second pool (in liters) = 4370 liters - 43 liters per minute * x minutes

(b) Now want to know when the two pools would have the same amount of water. If they have the same amount of water then you can express:

Amount of water in the first pool = Amount of water in the second pool

Replacing the expressions found in (a):

3850 liters - 33 liters per minute * x minutes= 4370 liters - 43 liters per minute * x minutes

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3 years ago

WARNJNG ⚠️⚠️ just help me please

marta [7]

(4)3 is also the same thing as 4x3 which is 12n

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3 years ago