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3 years ago

8

Blue whales eat an average of 5,000 pounds of fish daily, with a standard deviation of 850 pounds. Approximately what percentage

of blue whales eat more than 7,550 pounds of fish?
15.87%
5%
0.3%
0.15%

1 answer:

nikklg [1K]3 years ago

7 0

15.87% is the answer

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Help Asap! 20 points!

dybincka [34]

D = 10 is the diameter so r = d/2 = 10/2 = 5 is the radius

Plug r = 5 and h = 7 into the formula below and simplify
V = pi*r^2*h
V = pi*5^2*7
V = pi*25*7
V = pi*175
V = 175pi

Answer: Choice D

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3 years ago

ANTONII [103]

Answer:

It would be 15,00 overpriced

Step-by-step explanation:

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3 years ago

You place the spring vertically with one end on the floor. You then drop a book of mass 1.50 kg onto it from a height of 0.900 m

Alex777 [14]

Answer:

The value of the maximum distance the spring will be compressed

x = 0.143 m= 14.3 cm

Step-by-step explanation:

Given data

Mass m = 1.5 kg

h = 0.9 m

k = 1500 $\frac{N}{m}$

Since the book is released from a height h = 0.9 m above the top of the spring. so the total distance

y = h + x

Now in that case for the spring

$mg y= \frac{1}{2} k x^{2}$

$mg(h +x) = \frac{1}{2} k x^{2}$

By rearranging the above equation we get

$k x^{2} - 2mgx - 2mg h = 0$

Put all the values in above equation we get

$1500x^{2} - 29.43 x - 26.48 = 0$

By solving the above equation we get

x = 0.143 m= 14.3 cm

This is the value of the maximum distance the spring will be compressed.

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3 years ago

Simplify 10 to the power of 0

OverLord2011 [107]

Answer:

1

Step-by-step explanation:

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3 years ago

Read 2 more answers

(a) Find the equation of the normal to the hyperbola x = 4t, y = 4/t at the point (8,2).

gtnhenbr [62]

The slope of the tangent line to the curve at (8, 2) is given by the derivative $\frac{dy}{dx}$ at that point. By the chain rule,

$\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Differentiate the given parametric equations with respect to $t$ :

$x = 4t \implies \dfrac{dx}{dt} = 4$

$y = \dfrac4t \implies \dfrac{dy}{dt} = -\dfrac4{t^2}$

Then

$\dfrac{dy}{dx} = \dfrac{-\frac4{t^2}}4 = -\dfrac1{t^2}$

We have $x=8$ and $y=2$ when $t=2$ , so the slope at the given point is $\frac{dy}{dx} = -\frac14$ .

The normal line to the same point is perpendicular to the tangent line, so its slope is +4. Then using the point-slope formula for a line, the normal line has equation

$y - 2 = 4 (x - 8) \implies \boxed{y = 4x - 30}$

Alternatively, we can eliminate the parameter and express $y$ explicitly in terms of $x$ :

$x = 4t \implies t = \dfrac x4 \implies y = \dfrac4t = \dfrac4{\frac x4} = \dfrac{16}x$

Then the slope of the tangent line is

$\dfrac{dy}{dx} = -\dfrac{16}{x^2}$

At $x = 8$ , the slope is again $-\frac{16}{64}=-\frac14$ , so the normal has slope +4, and so on.

5 0

2 years ago