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3 years ago

6

The Stegosaurus dinosaur weighed about 2000 kilograms. The weight of his brain was only 0.004% of the weight of his body. How mu

ch did the Stegosaurus brain weigh?

1 answer:

Verizon [17]3 years ago

4 0

Answer:0.08 kilograms

Step-by-step explanation:

You might be interested in

Write the measure -128 30’ 45’’ as a decimal to the nearest thousandth

Zielflug [23.3K]

Answer:

-127.488°

Step-by-step explanation:

45" is 0.75 of 1', so -128° 30' 45" = -128° 30.75'

Next: 30.75' is 30.75/60 of 1°, and or -128° plus 0.5125 of 1°.

The final answer is found by adding 0.5125° to -128°:

-128.0000°

+ 0.5125°

--------------------

-127.4875

To the nearest thousandths of a degree, that would be -127.488°

6 0

3 years ago

Find the value of x, y, and z

kaheart [24]

Answer:

y=40

x=59

z=81

Step-by-step explanation:

therfore..40+59+81=180

5 0

3 years ago

Find the mean, median, mode, and range of this data: 49, 49, 54, 55, 52, 49, 55. If necessary, round to the nearest tenth.

4vir4ik [10]

The answer is A.

Hope I helped.

6 0

3 years ago

Read 2 more answers

26. If you know the lengths of two sides and the measure of _____ angle(s) of a right triangle, you can find the measure of the

lana [24]

Answer:

1

Step-by-step explanation:

For any triangle, you only need to know 3 pieces of information out of the 6 angles and sides to solve for the rest of the triangle.

4 0

3 years ago

A filtration process removes a random proportion of particulates in water to which it is applied. Suppose that a sample of water

Doss [256]

Answer:

$E(Y)=\frac{1}{25}$

Step-by-step explanation:

Let's start defining the random variables for this exercise :

$X_{1}:$ '' The proportion of the particulates that are removed by the first pass ''

$X_{2}:$ '' The proportion of what remains after the first pass that is removed by the second pass ''

$Y:$ '' The proportion of the original particulates that remain in the sample after two passes ''

We know the relation between the random variables :

$Y=(1-X_{1})(1-X_{2})$

We also assume that $X_{1}$ and $X_{2}$ are independent random variables with common pdf.

The probability density function for both variables is $f(x)=4x^{3}$ for $0$ and $f(x)=0$ otherwise.

The first step to solve this exercise is to find the expected value for $X_{1}$ and $X_{2}$ .

Because the variables have the same pdf we write :

$E(X_{1})= E(X_{2})=E(X)$

Using the pdf to calculate the expected value we write :

$E(X)=\int\limits^a_b {xf(x)} \, dx$

Where $a=$ ∞ and $b=$ - ∞ (because we integrate in the whole range of the random variable). In this case, we will integrate between $0$ and $1$ ⇒

Using the pdf we calculate the expected value :

$E(X)=\int\limits^1_0 {x4x^{3}} \, dx=\int\limits^1_0 {4x^{4}} \, dx=\frac{4}{5}$

⇒ $E(X)=E(X_{1})=E(X_{2})=\frac{4}{5}$

Now we need to use some expected value properties in the expression of $Y$ ⇒

$Y=(1-X_{1})(1-X_{2})$ ⇒

$Y=1-X_{2}-X_{1}+X_{1}X_{2}$

Applying the expected value properties (linearity and expected value of a constant) ⇒

$E(Y)=E(1)-E(X_{2})-E(X_{1})+E(X_{1}X_{2})$

Using that $X_{1}$ and $X_{2}$ have the same expected value $E(X)$ and given that $X_{1}$ and $X_{2}$ are independent random variables we can write $E(X_{1}X_{2})=E(X_{1})E(X_{2})$ ⇒

$E(Y)=E(1)-E(X)-E(X)+E(X_{1})E(X_{2})$ ⇒

$E(Y)=E(1)-2E(X)+[E(X)]^{2}$

Using the value of $E(X)$ calculated :

$E(Y)=1-2(\frac{4}{5})+(\frac{4}{5})^{2}=\frac{1}{25}$

$E(Y)=\frac{1}{25}$

We find that the expected value of the variable $Y$ is $E(Y)=\frac{1}{25}$

3 0

3 years ago