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2 years ago

This is for middle school 9th grade please help me :)

Mathematics

1 answer:

Stolb23 [73]2 years ago

7 0

Answer:

i think it's undefined

Step-by-step explanation:

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Percentage grade averages were taken across all disciplines at a particular university, and the mean average was found to be 83.

Georgia [21]

Correct Ans:
Option A. 0.0100

Solution:
We are to find the probability that the class average for 10 selected classes is greater than 90. This involves the utilization of standard normal distribution.

First step will be to convert the given score into z score for given mean, standard deviation and sample size and then use that z score to find the said probability. So converting the value to z score:

$z-score= \frac{90-83.6}{ \frac{8.7}{ \sqrt{10} } } \\ \\ 
z-score =2.326$

So, 90 converted to z score for given data is 2.326. Now using the z-table we are to find the probability of z score to be greater than 2.326. The probability comes out to be 0.01.

Therefore, there is a 0.01 probability of the class average to be greater than 90 for the 10 classes.

5 0

3 years ago

Can you please help me

shusha [124]

Answer:

1: inequality

2: solution

3: open circle

4: infinite

5: closed circle

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What is the power of 256

Juli2301 [7.4K]

In mathematics. 256 is a composite number, with the factorization 256 = 28, which makes it a power of two. 256 is 4 raised to the 4th power, so in tetration notation 256 is 24.

4 0

3 years ago

A trader bought 60 article for N25000 and sold them at N6000 per dozen.

Svet_ta [14]

Step-by-step explanation:

N6000×5=N30000

1)N30000-N25000=N5000

2)N5000÷N25000×100℅=20℅

3 0

2 years ago

Determine whether the set of vectors <img src="https://tex.z-dn.net/?f=%20v_%7B1%3D%283%2C2%2C1%29%2C%20v_%7B2%7D%20%3D%28-1%2C-

Korolek [52]

Since each vector is a member of $\mathbb R^3$ , the vectors will span $\mathbb R^3$ if they form a basis for $\mathbb R^3$ , which requires that they be linearly independent of one another.

To show this, you have to establish that the only linear combination of the three vectors $c_1\mathbf v_1+c_2\mathbf v_2+c_3\mathbf v_3$ that gives the zero vector $\mathbf0$ occurs for scalars $c_1=c_2=c_3=0$ .

$c_1\begin{bmatrix}3\\2\\1\end{bmatrix}+c_2\begin{bmatrix}-1\\-2\\-4\end{bmatrix}+c_3\begin{bmatrix}1\\1\\-1\end{bmatrix}=(0,0,0)\iff\begin{bmatrix}3&-1&1\\2&-2&1\\1&-4&-1\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$

Solving this, you'll find that $c_1=c_2=c_3=0$ , so the vectors are indeed linearly independent, thus forming a basis for $\mathbb R^3$ and therefore they must span $\mathbb R^3$ .

4 0

3 years ago