What expression is equivalent to 6 × 10^−4/2 × 10^3

Mathematics

1 answer:

Ksivusya [100]4 years ago

4 0

The correct answer is: [A]: " 3 * 10 ⁻⁷ " .

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Explanation:

To solve:

Given:

<u> (6 * 10⁻⁴ ) </u> = (6 ÷ 2) * 10⁽⁻⁴ ⁻³⁾ = 3 * 10 ⁻⁷ ; which is:

(2 * 10³ )

→ Answer choice: [A]: " 3 * 10 ⁻⁷ " .

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The following are the last 10 run scores Colin got in cricket:

Solnce55 [7]

Answers:

a) Mean = 20.4
b) New mean = 20

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Explanation:

To get the mean, we add up the scores and divide by 10 (because there are 10 scores at first)

28+13+4+12+32+22+13+22+26+32 = 204

204/10 = 20.4

The mean is 20.4

------------

For part b), we redo those steps shown above, but tack 16 onto the list. So we'll add up all the values (including that 16 at the end) and divide by 11 this time.

28+13+4+12+32+22+13+22+26+32+16 = 220

220/11 = 20

The new mean is 20.

The new mean is slightly smaller than the old mean. Notice how 16 is smaller than 20.4, so this new score pulls down the mean just a little bit.

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3 years ago

Write an inequality to represent the phrase: The sum of a number and 6 is no less than 31.

Nastasia [14]

Answer

6+x=31 0r 6+ x = <31

Step-by-step explanation:

3 0

3 years ago

A trucking company uses matrix A , shown below, to represent the driving time in hours for 6 trucking routes. A = [ 4 6 8 ]

zepelin [54]

Each element of the matrix are multiplied by the scalar to form a matrix of

same size as the original matrix in matrix scalar multiplication.

$60 \cdot A = \left[\begin{array}{ccc}240\4&360&480\\360&480&600\end{array}\right]$

Reasons:

The matrix <em>A</em> is presented as follows;

$A = {\left[\begin{array}{ccc}4&6&8\\6&8&10\end{array}\right]}$

Using the multiplication of a matrix and a scalar, we have;

$60 \cdot A = 60 \cdot \left[\begin{array}{ccc}4&6&8\\6&8&10\end{array}\right] = \left[\begin{array}{ccc}60 \times 4&60 \times 6&60 \times 8\\60 \times 6&60 \times 8&60 \times 10\end{array}\right] = \left[\begin{array}{ccc}\mathbf{240}&\mathbf{360}&\mathbf{480}\\\mathbf{360}&\mathbf{480}&\mathbf{600}\end{array}\right]$