I don’t understand how to do any or this.

The weight of an organ in adult males has a bell-shaped distribution with a mean of 320 grams and a standard deviation of 45 gr

irina1246 [14]

Answer:

a) 99.7%

b) 68%

c) 68% of organs weighs between 275 and 365 grams, so 32% of organs weighs will be less than 275 or more than 365 grams.

d) 81.5% of organs weighs between 290 grams and 365 grams.

Step-by-step explanation:

A) 99.7% of organs will be between 3 standard deviation from the mean.

$320 - 3 \times 45 = 185$

$320 + 3 \times 45 = 455$

So 99.7% of organs will be between 185 and 455.

B) 275 grams and 365 grams are 1 standard deviation from the mean.

From empirical rule about 68% data falls within 1 standard deviation from the mean.

So 68% of organs weighs btwn 275 grams and 365 grams.

C) Since 68% of organs weighs between 275 and 365 grams, so 32% of organs weighs will be less than 275 or more than 365 grams.

D) 230 is 2 standard deviation below the mean and 365 is one standard deviation above the mean.

According to the empirical rule, 95% of the observation lies within 2 standard deviations of the mean. Therefore 5% lies outside 2 standard deviations of the mean.

So 95%/2 = 47.5% of organs weighs between the mean and 230 grams.

According to the empirical rule about 68% data falls within one standard deviation from the mean.

so, 68%/2 = 34% of organs weighs between the mean and 365 grams.

So total = 47.5% + 34% = 81.5% of organs weighs between 290 grams and 365 grams.

7 0

3 years ago

Suppose that v is an eigenvector of matrix A with eigenvalue λA, and it is also an eigenvector of matrix B with eigenvalue λB. (

galben [10]

Answer:

(a) Yes, $λ_{A}+λ_{B}$

(b) Yes, $λ_{A}λ_{B}$

Step-by-step explanation:

First, lets understand what are eigenvectors and eigenvalues?

Note: I am using the notation $λ_{A}$ to denote Lambda(A) sign.

$v$ is an eigenvector of matrix A with eigenvalue $λ_{A}$

$v$ is also eigenvector of matrix B with eigenvalue $λ_{B}$

So we can write this in equation form as

$Av=λ_{A}v$

So what does this equation say?

When you multiply any vector by A they do change their direction. any vector that is in the same direction as of $Av$ , then this $v$ is called the eigenvector of $A$ . $Av$ is $λ_{A}$ times the original $v$ . The number $λ_{A}$ is the eigenvalue of A.

$λ_{A}$ this number is very important and tells us what is happening when we multiply $Av$ . Is it shrinking or expanding or reversed or something else?

It tells us everything we need to know!

Bonus:

By the way you can find out the eigenvalue of $Av$ by using the following equation:

$det(A-λI)=0$

where I is identity matrix of the size of same as A.

Now lets come to the solution!

(a) Show that $v$ is an eigenvector of $A + B$ and find its associated eigenvalue.

The eigenvalues of $A$ and $B$ are $λ_{A}$ and $λ_{B}$ , then

$(A+B)(v)=Av+Bv=(λ_{A})v + (λ_{B})v=(λ_{A}+λ_{B})(v)$

so, $(A+B)(v)=(λ_{A}+λ_{B})(v)$

which means that $v$ is also an eigenvector of $A+B$ and the associated eigenvalues are $λ_{A}+λ_{B}$

(b) Show that $v$ is an eigenvector of $AB$ and find its associated eigenvalue.

The eigenvalues of $A$ and $B$ are $λ_{A}$ and $λ_{B}$ , then

$(AB)(v)=A(Bv)=A(λ_{B})=λ_{B}(Av)=λ_{B}λ_{A}(v)=λ_{A}λ_{B}(v)$

so,

$(AB)(v)=λ_{A}λ_{B}(v)$

which means that $v$ is also an eigenvector of $AB$ and the associated eigenvalues are $λ_{A}λ_{B}$

5 0

2 years ago

What is one hundred forty-one million,six hundred twenty thousand in standard form

Digiron [165]

Millions thousands hundreds
141, 620, 000.

6 0

3 years ago

Read 2 more answers

4.

svet-max [94.6K]

Buy 5 for $12 and buy 4 for $15

8 0

3 years ago

What is the HCF of two consecutive numbers?

Luba_88 [7]

Answer: The HCF o two consecutive numbers is always one.

Step-by-step explanation: The reason behind this is that the two consecutive numbers do not have any common factor other than 1. Hence 1 becomes the highest common factor between two consecutive numbers

please mark brainliest

6 0

2 years ago