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3 years ago

What is 48,100,000,000,000 in scientific notation?. . A.4.8110^-11. . B.4.8110^-13. . C.4.8110^11. . D.4.8110^13

Mathematics

2 answers:

m_a_m_a [10]3 years ago

5 0

D. 4.81 x 10^13 is the answer hope it helps ;)

Sindrei [870]3 years ago

3 0

D. 4.81 x 10^13
Move the decimal to the right 13 times, because when you multiply by 10, you make the number bigger, when you divide you move the decimal to the left to make the number smaller.
48,100,000,000,000

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Step-by-step explanation:

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The side of a square measures (3x − 6) units. Part A: What is the expression that represents the area of the square? Show your w

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A. The area of a square is given as:

A = s^2

Where s is a measure of a side of a square. s = (2 x – 5) therefore,

A = (2 x – 5)^2

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A = 4 x^2 – 20 x + 25

B. The degree of a polynomial is the highest exponent of the variable x, in this case 2. Therefore the expression obtained in part A is of 2nd degree.

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3 years ago

Mystery Boxes: Breakout Rooms

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Answer:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

Required

Fill in the box

From the question, the range is:

$Range = 60$

Range is calculated as:

$Range = Highest - Least$

From the box, we have:

$Least = 1$

So:

$60 = Highest - 1$

$Highest = 60 +1$

$Highest = 61$

The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

$IQR = 20$ --- interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

$Q_3$ is the median of the upper half while $Q_1$ is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}$

Upper half

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

The quartile is calculated by calculating the median for each of the above halves is calculated as:

$Median = \frac{N + 1}{2}th$

Where N = 7

So, we have:

$Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th$

So,

$Q_3$ = 4th item of the upper halves

$Q_1$ = 4th item of the lower halves

From the upper halves

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We have:

$Q_3 = 32$

$Q_1$ can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

So:

$20 = 32 - Q_1$

$Q_1 = 32 - 20$

$Q_1 = 12$

So, the lower half becomes:

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}$

From this, the updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question, the median is:

$Median = 22$ and $N = 14$

To calculate the median, we make use of:

$Median = \frac{N + 1}{2}th$

$Median = \frac{14 + 1}{2}th$

$Median = \frac{15}{2}th$

$Median = 7.5th$

This means that, the median is the average of the 7th and 8th items.

The 7th and 8th items are blanks.

However, from the question; the mode is:

$Mode = 18$

Since the values of the box are in increasing order and the average of 18 and 18 do not equal 22 (i.e. the median), then the 7th item is:

$7th = 18$

The 8th item is calculated as thus:

$Median = \frac{1}{2}(7th + 8th)$

$22= \frac{1}{2}(18 + 8th)$

Multiply through by 2

$44 = 18 + 8th$

$8th = 44 - 18$

$8th = 26$

The updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question.

$Mean = 26$

Mean is calculated as:

$Mean = \frac{\sum x}{n}$

So, we have:

$26= \frac{1 + 2nd + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 12th + 58 + 61}{14}$

Collect like terms

$26= \frac{ 2nd + 12th+1 + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 58 + 61}{14}$

$26= \frac{ 2nd + 12th+304}{14}$

Multiply through by 14

$14 * 26= 2nd + 12th+304$

$364= 2nd + 12th+304$

This gives:

$2nd + 12th = 364 - 304$

$2nd + 12th = 60$

From the updated box,

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We know that:

The 2nd value can only be either 2 or 3

The 12th value can take any of the range 33 to 57

Of these values, the only possible values of 2nd and 12th that give a sum of 60 are:

$2nd = 3$

$12th = 57$

i.e.

$2nd + 12th = 60$

$3 + 57 = 60$

So, the complete box is:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

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2 years ago

6. April works as an emergency medical technician (EMT) and is a photographer. As an EMT she earns $19 per

ElenaW [278]

Answer: 19t +100

Step-by-step explanation:

19t +100 where t is the number of hours she worked and the extra 100 dollars is the money she got from her friend.

6 0

3 years ago

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In the figure below, GRAH is a rectangle, RA=7, and GA=10. Determine AH.

svetoff [14.1K]

Answer:

AH = $\sqrt{51}$

Step-by-step explanation:

The opposite sides of a rectangle are congruent , so

GH = RA = 7

Using Pythagoras' identity in right Δ GHA

AH² + GH² = GA²

AH² + 7² = 10²

AH² + 49 = 100 ( subtract 49 from both sides )

AH² = 51 ( take the square root of both sides )

AH = $\sqrt{51}$ ≈ 7.14 ( to 2 dec. places )

3 0

3 years ago

What is 48,100,000,000,000 in scientific notation?. . A.4.81*10^-11. . B.4.81*10^-13. . C.4.81*10^11. . D.4.81*10^13

What is 48,100,000,000,000 in scientific notation?. . A.4.8110^-11. . B.4.8110^-13. . C.4.8110^11. . D.4.8110^13