Find the coefficient of<br>a^3

The ages of trees in a forest are normally distributed with a mean of 25 years and a standard deviation of 4. Approximately what

harkovskaia [24]

Answer:

78.88%

Step-by-step explanation:

We have been given that

$\mu=25,\sigma=4,x_1=20,x_2=30$

The z-score formula is given by

$z-\text{score}=\frac{x-\mu}{\sigma}$

For $x_1=20$

$z_1=\frac{20-25}{4}\\\\z_1=-1.25$

For $x_2=30$

$z_2=\frac{30-25}{4}\\\\z_2=1.25$

Now, we find the corresponding probability from the standard z score table.

For the z score -1.25, we have the probability 0.1056

For the z score 1.25, we have the probability 0.8944

Therefore, the percent of the trees that are between 20 and 30 years old is given by

0.8944 - 0.1056

= 0.7888

=78.88%

6 0

3 years ago

Read 2 more answers

Consider the linear transformation T from V = P2 to W = P2 given by T(a0 + a1t + a2t2) = (2a0 + 3a1 + 3a2) + (6a0 + 4a1 + 4a2)t

Svet_ta [14]

Answer:

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

Step-by-step explanation:

First we start by finding the dimension of the matrix [T]EE

The dimension is : Dim (W) x Dim (V) = 3 x 3

Because the dimension of P2 is the number of vectors in any basis of P2 and that number is 3

Then, we are looking for a 3 x 3 matrix.

To find [T]EE we must transform the vectors of the basis E and then that result express it in terms of basis E using coordinates and putting them into columns. The order in which we transform the vectors of basis E is very important.

The first vector of basis E is e1(t) = 1

We calculate T[e1(t)] = T(1)

In the equation : 1 = a0

$T(1)=(2.1+3.0+3.0)+(6.1+4.0+4.0)t+(-2.1+3.0+4.0)t^{2}=2+6t-2t^{2}$

$[T(e1)]E=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

And that is the first column of [T]EE

The second vector of basis E is e2(t) = t

We calculate T[e2(t)] = T(t)

in the equation : 1 = a1

$T(t)=(2.0+3.1+3.0)+(6.0+4.1+4.0)t+(-2.0+3.1+4.0)t^{2}=3+4t+3t^{2}$

$[T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right]$

Finally, the third vector of basis E is $e3(t)=t^{2}$

$T[e3(t)]=T(t^{2})$

in the equation : a2 = 1

$T(t^{2})=(2.0+3.0+3.1)+(6.0+4.0+4.1)t+(-2.0+3.0+4.1)t^{2}=3+4t+4t^{2}$

Then

$[T(t^{2})]E=\left[\begin{array}{c}3&4&4\\\end{array}\right]$

And that is the third column of [T]EE

Let's write our matrix

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

T(X) = AX

Where T(X) is to apply the transformation T to a vector of P2,A is the matrix [T]EE and X is the vector of coordinates in basis E of a vector from P2

For example, if X is the vector of coordinates from e1(t) = 1

$X=\left[\begin{array}{c}1&0&0\\\end{array}\right]$

$AX=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]\left[\begin{array}{c}1&0&0\\\end{array}\right]=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

Applying the coordinates 2,6 and -2 to the basis E we obtain

$2+6t-2t^{2}$

That was the original result of T[e1(t)]

8 0

3 years ago

Please help me. This is due today

sattari [20]

Answer:

it answer is 3 because it is trigonometry

3 0

3 years ago

HELP ME WITH THESE TWO PLS !! I'LL GIVE BRAINLIEST

GenaCL600 [577]

Answer:

that is a two sided triangle

6 0

3 years ago

An ambitious statistical-education researcher wants to analyze the typical statistical skills of STEM (science, technology, engi

antiseptic1488 [7]

Answer:

Option b, c and e are wonderful approaches to solve the problem.

Step-by-step explanation:

Option (b) is appropriate this is because the option is talking about Simple random sampling where random universities are chosen to remove bias.

Option (c) is correct because this is an example of Stratified sampling where two homogenous groups (private and public universities are considered) and samples are chosen at random to remove bias

Option (e) is correct because this again is an example of Simple random sampling where 60 random STEM majors are chosen at random.

7 0

3 years ago

Find the coefficient ofa^3​

Find the coefficient ofa^3