Ask question

Ask question

All categories

2 years ago

6

Given a normally distributed data set with the mean = 88 and the standard deviation = 6, according to the empirical rule, what p

ercent of the data points are above 76?

1 answer:

vladimir2022 [97]2 years ago

3 0

According to the empirical rule, the percent of the data points that are above 76 is; 2.28%

<h3>How to use Empirical Rule in Statistics?</h3>

We are given;

Mean; μ = 88

Standard deviation; σ = 6

Sample mean; x' = 76

Formula for z-score is;

z = (x' - μ)/σ

z = (76 - 88)/6

z = -2

Now, from the empirical rule of normal distribution graph attached, we see that at z = -2, the percentage is 95.44%. However, we want to find the percentage that lie above 27. Thus, this is (100% - 95.44%)/2 = 2.28%

Read more about Statistics Empirical Rule at; brainly.com/question/10093236

#SPJ1

You might be interested in

Does the equation 2x + 3 = 11 go through the point (1,4)?

Jet001 [13]

Answer: yes it does

2x= 8
x= 8/2

x=4.

6 0

3 years ago

Mrs. Pascal has a candy jar that contains 5 Jolly Ranchers and 13 jawbreakers. Solve each of the following problems and explain

ZanzabumX [31]

Answer:

260 jawbreakers

Step-by-step explanation:

The ratio is 5:13 so if the value of Jolly ranchers increased 20x then the number of jawbreakers would also increase by 20x. There are 260 jawbreakers.

8 0

3 years ago

Use the interactive number line to find each sum to complete the table

Likurg_2 [28]

-1 - 2 = -3

R = -3

-4 + 1 = -3

S = -3

-6 - 3 = -9

T = -9

6 0

3 years ago

Read 2 more answers

An electronics company produces transistors, resistors, and computer chips. Each transistor requires 3 units of copper, 1 unit

padilas [110]

Answer:

An electronics company can be produce 350 transistors and 340 computer chips, they can´t produce resistors.

Step-by-step explanation:

1. We will name the variables for transistors, resistors and the computer chips.

a = Transistors

b= Resistors

c = Computer chips

2. We propose three linear equations, one for the copper, one for the zinc and one for the glass.

$\left \{ {{3a+3b+2c=1730} \atop {a+2b+c=690}}\atop {2a+b+2c=1380}} \right.$

3. We write the matrix form as Ax=d

$A=\left(\begin{array}{ccc}3&3&2\\1&2&1\\2&1&2\end{array}\right)$

$x=\left(\begin{array}{ccc}a\\b\\c\end{array}\right)$

$A=\left(\begin{array}{ccc}1730\\690\\1380\end{array}\right)$

With this formula the solution of x is $x=\frac{d}{A}$ or $x=A^{-1}d$

4. We will find the inverse matrix $A^{-1}$ using the formula:

$A^{-1} = \frac{1}{detA} (C_{A})^{T}$

a. det A

$det A=\left[\begin{array}{ccc}3&3&2\\1&2&1\\2&1&2\end{array}\right] =3*(4-1)-3*(2-2)+2*(1-4)=9-0-6=3$

b. $(C_{A})^{T}$

$C_{A}=\left(\begin{array}{ccc}4-1&.(2-2)&1-4\\-(6-2)&6-4&-(3-6)\\3-4&-(3-2)&6-3\end{array}\right)$

$C_{A}=\left(\begin{array}{ccc}3&.0&-3\\-4&2&3\\-1&-1&3\end{array}\right)$

$(C_{A}) ^T=\left(\begin{array}{ccc}3&0&-3\\-4&2&3\\-1&-1&3\end{array}\right)^T$

$(C_{A}) ^T=\left(\begin{array}{ccc}3&-4&-1\\0&2&-1\\-3&3&3\end{array}\right)$

c. $A^{-1}$

$A^{-1}=\frac{1}{3} \left(\begin{array}{ccc}3&-4&-1\\0&2&-1\\-3&3&3\end{array}\right)$

5. As $x=\frac{d}{A}$ or $x=A^{-1}d$ , the solution of x is:

$x=\frac{1}{3}\left(\begin{array}{ccc}3&-4&-1\\0&2&-1\\-3&3&3\end{array}\right)\left(\begin{array}{ccc}1730\\690\\1380\end{array}\right)$

$x=\frac{1}{3}\left(\begin{array}{ccc}(3*1730)+(-4*690)+(-1*1380)\\(0*1730)+(2*690)+(-1*1380)\\(-3*1730)+(3*690)+(3*1380)\end{array}\right)$

$x=\frac{1}{3}\left(\begin{array}{ccc}1050\\0\\1020)\end{array}\right)$

$X=\left[\begin{array}{ccc}350\\0\\340\end{array}\right]$

<u><em>Therefore:</em></u>

<u><em>a= 350 Transistors</em></u>

<u><em>b=0 Resistors</em></u>

<u><em>c=340 Computer chips</em></u>

4 0

3 years ago

Find the 63rd term for the sequence,<br> 8, 11, 14,...

Mariana [72]

Answer:

194

Step-by-step explanation:

t63 = t1 + (n - 1)(d)

= 8 + (63 - 1)(3)

= 8 + (62)(3)

= 8 + 186

= 194

6 0

2 years ago