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Agata [3.3K]
3 years ago
10

A set of data whose histogram is extremely skewed yields a sample mean and standard deviation of 68.5 and 12.75, respectively. W

hat is the minimum percentage of observations that:
A. are between 43 and 94.

Percentage = At Least %

B. are between 30.25 and 106.75.

Percentage = At Least %
Mathematics
1 answer:
Montano1993 [528]3 years ago
8 0


any other questions or comments about this.
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Simplify the following expressions. All answers should be in standard form.
Ratling [72]

Answer:

The given expression in simplified form is: 15x^2+10x

Step-by-step explanation:

Simplifying the expression involve reducing the number of terms to minimum and making the expressions in standard form.

This may involve addition, subtraction and multiplication of variables.

Given expression is:

5x(3x+2)

Distributing 5x will give us:

= 5.3.x.x + 5.2.x\\=15x^2+10x

Hence,

The given expression in simplified form is: 15x^2+10x

6 0
3 years ago
I need help with this homework can anyone help me. I need work too, help is appreciated:)
quester [9]
Let x = # of tickets sold in advance 
Let y = # of tickets sold the day of 


Cost of the tickets & total sales: 6x & 10y = 6828

You also know y = x + 206

Take the equation mentioned above y = x + 206 and sub it in anywhere the variable y is in the other equations so you'll have this:
6x + 10(x+206) = 6828

Now solve for x to get x = 298

To finish the problem, you must now find the number of y tickets sold.
Sub your x value that you found back into the equation y = x + 206 and you'll get y = 504.

So, 298 tickets were sold in advance and 504 tickets were sold the day of
3 0
3 years ago
Read 2 more answers
Which of the following gives all of the sets that contain √9?
romanna [79]

Soldiers for the nation of Gilead. “Blessed Be the Fruit”: The standard greeting amongst Gilead residents. The traditional reply is “May the Lord open.” “Under His Eye”: The Gileadean equivalent of “Aloha” — it works as both a hello and a goodbye.

7 0
2 years ago
The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes
saul85 [17]

Answer:

a)0.08  , b)0.4  , C) i)0.84  , ii)0.56

Step-by-step explanation:

Given data

P(A) =  professor arrives on time

P(A) = 0.8

P(B) =  Student aarive on time

P(B) = 0.6

According to the question A & B are Independent  

P(A∩B) = P(A) . P(B)

Therefore  

{A}' & {B}' is also independent

{A}' = 1-0.8 = 0.2

{B}' = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B')  (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

P(\frac{B'}{A}) = \frac{P(B'\cap A)}{P(A)} = \frac{0.4\times 0.8}{0.8} = 0.4

Part c)

Assume the events are not independent

Given Data

P(\frac{{A}'}{{B}'}) = 0.4

=\frac{P({A}'\cap {B}')}{P({B}')} = 0.4

P({A}'\cap {B}') = 0.4 x P({B}')

= 0.4 x 0.4 = 0.16

P({A}'\cap {B}') = 0.16

i)

The probability that at least one of them is on time

P(A\cup B) = 1- P({A}'\cap {B}')  

=  1 - 0.16 = 0.84

ii)The probability that they are both on time

P(A\cap  B) = 1 - P({A}'\cup {B}') = 1 - [P({A}')+P({B}') - P({A}'\cap {B}')]

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0
3 years ago
Henry has 15 coins in his pocket. 10 of the coins are pennies, 5 of the coins are quarters. Henry pulls one coin from his pocket
Nadya [2.5K]

Answer:

The probility is hight .

Step-by-step explanation:

because there are more pennies then quarters so there's about a 85 to 75% chance

4 0
2 years ago
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