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IrinaVladis [17]
3 years ago
7

Evaluate (One-fourth) cubed. Check all that apply. The base is One-fourth. The expanded form is (One-fourth) (one-fourth) (one-f

ourth). The expanded form is StartFraction 1 times 1 times 1 Over 4 EndFraction. (One-fourth) cubed = one-fourth (One-fourth) cubed = StartFraction 1 Over 64 EndFraction
Mathematics
2 answers:
Vedmedyk [2.9K]3 years ago
8 0

Answer:

The base is One-fourth.

The expanded form is (One-fourth) (one-fourth) (one-fourth).

(One-fourth) cubed = StartFraction 1 Over 64 EndFraction

Step-by-step explanation:

Given:

Evaluate (One-fourth) cubed.

(1/4)^3

= (1/4) (1/4) (1/4)

= 1/64

The base is One-fourth.

The expanded form is (One-fourth) (one-fourth) (one-fourth).

(One-fourth) cubed = StartFraction 1 Over 64 EndFraction

FrozenT [24]3 years ago
4 0

Answer:

its 1, 2, 5.

Step-by-step explanation:

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3 years ago
There are 10 employees in a particular division of a company. Their salaries have a mean of 570,000, a median of $55,000,and a s
harkovskaia [24]

Answer:

a) $160,000

b) $55,000

c) $332264.804

Step-by-step explanation:

We are given that there are 10 employees in a particular division of a company and their salaries have a mean of $70,000, a median of $55,000, and a standard deviation of $20,000.

And also the largest number on the list is $100,000 but By accident, this number is changed to $1,000,000.

a) Value of mean after the change in value is given by;

     Original Mean = $70,000

       \frac{\sum X}{n} = $70,000  ⇒ \sum X = 70,000 * 10 = $700,000

   New \sum X after change = $700,000 - $100,000 + $1,000,000 = $1600000

  Therefore, New mean = \frac{1600000}{10} = $160,000 .

b) Median will not get affected as median is the middle most value in the data set and since $1,000,000 is considered to be an outlier so median remain unchanged at $55,000 .

c) Original Variance = 20000^{2} i.e.  20000^{2} = \frac{\sum x^{2} - n*xbar }{n -1}

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    New Variance = \frac{new\sum x^{2} - n*new xbar }{n -1} = \frac{9.936007 *10^{11}  - 10*160000 }{10 -1} = 1.103999 * 10^{11}    Therefore, standard deviation after change = \sqrt{1.103999 * 10^{11} } = $332264.804 .

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3 years ago
A sample of marble has a volume of 6 cm3 and a density of 2.76 g/cm3. What is its mass?
Drupady [299]

Answer:

16.56\ g

Step-by-step explanation:

we know that

The density is equal to the mass divided by the volume

Let

x-----> the mass of marble

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substitute and solve for x

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3 0
3 years ago
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