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3 years ago

Which of the following estimates at a 95% confidence level most likely comes from a small sample?

Mathematics

2 answers:

horsena [70]3 years ago

7 0

It might be B but double check!

earnstyle [38]3 years ago

6 0

I think it’s B but I don’t really know so…

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A scientist inoculates mice, one at a time, with a disease germ until he finds 3 that have contracted the disease. If the prob

UkoKoshka [18]

Answer:

The probability that 8 mice are required is 0.2428.

Step-by-step explanation:

Given : A scientist inoculates mice, one at a time, with a disease germ until he finds 3 that have contracted the disease. If the probability of contracting the disease is two sevenths.

To find : What is the probability that 8 mice are required? The probability that that 8 mice are required is nothing ?

Solution :

Applying binomial distribution,

$P(X=r)=^nC_r p^rq^{n-r}$

Where, p is the probability of success $p=\frac{2}{7}$

q is the probability of failure q=1-p, $q=1-\frac{2}{7}=\frac{5}{7}$

n is total number of trials n=8

r=3

Substitute the values,

$P(X=3)=^8C_3 (\frac{2}{7})^3 (\frac{5}{7})^{8-3}$

$P(X=3)=\frac{8!}{3!5!}\times \frac{8}{343}\times (\frac{5}{7})^{5}$

$P(X=3)=\frac{8\times 7\times 6}{3\times 2\times 1}\times \frac{8}{343}\times \frac{3125}{16807}$

$P(X=3)=0.2428$

Therefore, the probability that 8 mice are required is 0.2428.

5 0

3 years ago

2 1/7 - 1/14 =<br><br>I rlly don't understand

lutik1710 [3]

Answer:

2 1/14

Step-by-step explanation:

First, we will equalize the denominators of the fractions.

(1/7) × 2 = 2/14

Then we can solve this equation.

2 2/14 - 1/14 = 2 1/14

6 0

2 years ago

Solve the given system of equations using either Gaussian or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTI

cricket20 [7]

Answer:

The system has infinitely many solutions

$\begin{array}{ccc}x_1&=&-x_3\\x_2&=&-x_3\\x_3&=&arbitrary\end{array}$

Step-by-step explanation:

Gauss–Jordan elimination is a method of solving a linear system of equations. This is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.

An Augmented matrix, each row represents one equation in the system and each column represents a variable or the constant terms.

There are three elementary matrix row operations:

Switch any two rows
Multiply a row by a nonzero constant
Add one row to another

To solve the following system

$\begin{array}{ccccc}x_1&-3x_2&-2x_3&=&0\\-x_1&2x_2&x_3&=&0\\2x_1&+3x_2&+5x_3&=&0\end{array}$

Step 1: Transform the augmented matrix to the reduced row echelon form

$\left[ \begin{array}{cccc} 1 & -3 & -2 & 0 \\\\ -1 & 2 & 1 & 0 \\\\ 2 & 3 & 5 & 0 \end{array} \right]$

This matrix can be transformed by a sequence of elementary row operations

Row Operation 1: add 1 times the 1st row to the 2nd row

Row Operation 2: add -2 times the 1st row to the 3rd row

Row Operation 3: multiply the 2nd row by -1

Row Operation 4: add -9 times the 2nd row to the 3rd row

Row Operation 5: add 3 times the 2nd row to the 1st row

to the matrix

$\left[ \begin{array}{cccc} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 0 \\\\ 0 & 0 & 0 & 0 \end{array} \right]$

The reduced row echelon form of the augmented matrix is

$\left[ \begin{array}{cccc} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 0 \\\\ 0 & 0 & 0 & 0 \end{array} \right]$

which corresponds to the system

$\begin{array}{ccccc}x_1&&-x_3&=&0\\&x_2&+x_3&=&0\\&&0&=&0\end{array}$

The system has infinitely many solutions.

$\begin{array}{ccc}x_1&=&-x_3\\x_2&=&-x_3\\x_3&=&arbitrary\end{array}$

7 0

3 years ago

6(m +2 + 7m <br> Please help please help

NISA [10]

Answer:

i think that is 2m + m14

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

URGENT!! Really need help on this test

Katyanochek1 [597]

Answer:

G= L &M

H=L &M

K=P&N

J=P&N

7 0

3 years ago