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Vikki [24]
3 years ago
5

Perform the indicated operation. 4/11 · 10/8 4/11 5/11 6/11

Mathematics
1 answer:
Dimas [21]3 years ago
7 0
4/11 * 10/8

Multiplying fractions
1) multiply the numerators: 4 * 10 = 40
2) multiply the denominators: 11 * 8 = 88
3) simplify the fraction
40/88 ⇒ 10/22 ⇒ 5/11

40 ÷ 4 = 10
88 ÷ 4 = 22

10 ÷ 2 = 5
22 ÷ 2 = 11

4/11 * 10/8 = 5/11
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Please help me it’s due today
Alchen [17]

Answer:

Northside

Median: 36

Range: 32

IQR: 17

Southside:

Median: 39

Range: 38

IQR: 22

Step-by-step explanation:

Northside first

Order the ages from least to greatest

24, 28, 32, 32, 32, 36, 41, 44, 49, 49, 56

Median is 36 since it's in the middle

Subtract the greatest number from the smallest number to find the range (56-24) = 32

To find the IQR you need the ages least to greatest

24, 28, 32, 32, 32, 36, 41, 44, 49, 49, 56

Find the lower quarter = 32

Find the upper quarter = 49

Subtract 49 and 32 (49-32) = 17

Southside

The Median is the line at the middle of the box, in this case its between 38 and 40 which is 39.

The range is 61 - 23 = 38 (Greatest - least)

The IQR is 52 - 30 = 22

Hope this helps!

7 0
2 years ago
Read 2 more answers
True or false: a mosaic plot is useful for visualizing the relationship between a numerical and a categorical variable.
Ivahew [28]

It is false that for visualizing the relationship between a numerical and a categorical variable, a mosaic plot is useful. This is because the best plot to use for this kind of relationship is the side by side box plot.

 

Comparing the distribution of a numerical variable across the levels of a categorical variable is something we usually consider in this type of relationship.

 

The levels of one categorical variable by means of a quantitative variable is best compared by using the visual display called the side by side boxplot. <span>By placing single boxplots adjacent to one another on a single scale, the side by side box plot is constructed.</span>

3 0
4 years ago
From past experience, a company has found that in carton of transistors: 92% contain no defective transistors 3% contain one def
jasenka [17]

Answer:

E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500

In order to find the variance we need to find first the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300

The variance is calculated with this formula:

Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075

And the standard deviation is just the square root of the variance and we got:

Sd(X) = \sqrt{0.3075}= 0.5545

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

Solution to the problem

LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X

X         0           1           2         3

P(X)    0.92     0.03    0.03     0.02

We can calculate the expected value with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i)

And replacing we got:

E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500

In order to find the variance we need to find first the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300

The variance is calculated with this formula:

Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075

And the standard deviation is just the square root of the variance and we got:

Sd(X) = \sqrt{0.3075}= 0.5545

8 0
4 years ago
The library is having a book sale. Hardcover books sell for $4 each, and paperback books are $2 each. If Ayanna spends $26 for 8
Liula [17]

Answer: 6

Step-by-step explanation: I divided $26 by $4

5 0
3 years ago
Read 2 more answers
Iodine-131 is a radioactive material with half-life of eight days. A scientist starts with 100 grams of iodinie- 131. Which func
MrRissso [65]
The correct answer is Choice A.

This is an example of an exponential equation, so we need the formula y=ab^x.

The a value is the starting value of 100. The b value must be a decimal lower than 1 because it is decreases.

If you substitute in 8 for x, you will see that the output is about 50 (half of 100).
6 0
4 years ago
Read 2 more answers
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