1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Salsk061 [2.6K]
2 years ago
6

Fawzi's SUV is 6 feet 4 inches

Mathematics
1 answer:
Nat2105 [25]2 years ago
8 0

Answer:

<u>9 feet 2 inches</u>

Step-by-step explanation:

Given :

⇒ Height of SUV = <u>6 feet 4 inches</u>

⇒ Height of box = <u>2 feet 10 inches</u>

<u />

============================================================

Solving :

⇒ Total Height = Height of SUV + Height of box

⇒ Total Height = 6 feet 4 inches + 2 feet 10 inches

⇒ Total Height = 8 feet 14 inches

* But, 12 inches = 1 feet *

Hence,

⇒ Total Height = 8 feet + 1 foot + 2 inches

⇒ Total Height = <u>9 feet 2 inches</u>

You might be interested in
If i had 7 apples and someone took away 5 how many would i have left
Alenkasestr [34]
You'll have no apples left
6 0
3 years ago
Read 2 more answers
Installments
ioda

Answer:

$ 338.70

Step-by-step explanation:

https://www.calculatorsoup.com/calculators/financial/loan-calculator-simple.php?pv=15%2C000.00&ratepercent=4&t=48&time_t=month&given_data_last=find_pmt&action=solve

6 0
3 years ago
Convert: 4 quarts (dry) to liters.<br> a. 0.4406<br> b. 4.406<br> c. 44.06<br> d. 440.6
kipiarov [429]
1 dry quart=1.10 

1.10 x 4= 4.4 

Final answer: B.  
4 0
3 years ago
Determine the point of intersection of right bisectors in a triangle ∆ with vertices A (-3, 5), B (1, 1) and (−7, −3). Find the
iris [78.8K]

Answer:

Point of intersection (-11/3 , 1/3)

All distances from the vertices are \frac{10\sqrt{2} }{3}

Step-by-step explanation:

The vertices of the triangle is A(-3,5) , B (1,1) and C (-7,-3)

We need to find the perpendicular bisector of the triangle first

let's take one side connecting A and B

mid point of A and B = (-1,3)

Slope of the line joining A and B = -1

slope of perpendicular to line joining A and B = 1

equation of line passing through (-1,3) with slope 1

y - 3 = 1(x-(-1))

y -3 = x+1

x-y = -4 ............(1)

similarly

mid point joining B and C = (-3,-1)

slope perpendicular to line joining B and C  =  -2

Equation of perpendicular bisector of line joining B and C =

y +1 = -2(x +3 )

y+1 = -2x -6

2x+y = -7 ..........(2)

On solving 1 and 2

x= -11/3 , y= 1/3

Distances

From A = \frac{10\sqrt{2} }{3}

From B = \frac{10\sqrt{2} }{3}

From C = \frac{10\sqrt{2} }{3}

7 0
3 years ago
Mystery Boxes: Breakout Rooms
ollegr [7]

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)

<u>Lower half:</u>

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

<u>Upper half</u>

<u></u>\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}<u></u>

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

<u></u>\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}<u></u>

<u></u>

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:

<u>Lower half:</u>

\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:

<em>The 2nd value can only be either 2 or 3</em>

<em>The 12th value can take any of the range 33 to 57</em>

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}

6 0
2 years ago
Other questions:
  • 2.8 division 14 that it the
    15·1 answer
  • 2x + 3y = 9
    13·2 answers
  • I want to know how to get the answer which is “3/10” can anyone tell me the steps to solving this probability problem?
    6·1 answer
  • A marching band stands in a triangle formation. The drum major stands at A, one of the vertices of the triangle. The whole forma
    10·1 answer
  • When a chip fabrication facility is operating normally, the lifetime of a microchip operated at temperature T, measured in degre
    14·1 answer
  • Fatima evaluated the expression StartFraction 4 m Superscript negative 3 Baseline n Superscript negative 2 Baseline Over m Super
    15·2 answers
  • Answer a and b. Help!!!!
    11·1 answer
  • In a three digit number, the tens digit is equal to the sum of the hundreds digit and the units digit. The hundreds digit of thi
    7·2 answers
  • For an equation that represents as a function of the set of all values taken on by the ________ variable is the domain, and the
    11·1 answer
  • A. P is equal to the set containing t,v,c and d
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!