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Tresset [83]
3 years ago
5

Divide. Write your answer as a fraction or mixed number in simplest form. 9 1/3 • 7

Mathematics
1 answer:
nordsb [41]3 years ago
4 0

Answer:

65 1/3 or 196/3

Step-by-step explanation:

9 1/3 can be written as 28/3 (9 times 3 plus 1)

28/3 times 7/1 (because any whole number can have a 1 as the denominator)

Multiply top to top and bottom to bottom

28 times 7/ 3 times 1

196/3

Simplify to

65 1/3 but 196/3 will work out as well

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What is the equation of a line that passes through (1,5) and (3,11)? Show work please.
jekas [21]
First, find the gradient/slope:
Use slope formula:
m=(y2-y1)/(x2-x1)
=(11-5)/(3-1)
=6/2
=3
Then use the line equation formula:
y=mx+c
You can substitute (1,5) if you like, also must substitute the slope as well!
5=3x1+c
c=2
Then find the full equation, which gives you the answer:
y=3x+2
3 0
3 years ago
16. Beginning at 8:30 A.M., tours of the National Capitol and the White House begin at a tour agency. Tours for the National Cap
vazorg [7]
It would be every 60 minutes I believe!
7 0
3 years ago
A sandbox has an area of 10 2/3 square feet, and the length is 1/4 feet. What is the width of the sandbox?
AlekseyPX

Answer:

The width of the sandbox is 42\frac{2}{3} \ ft.

Step-by-step explanation:

Given,

Area of the sandbox = 10\frac{2}{3}=\frac{32}{3}\ ft^2

Length = \frac{1}{4}\ ft

Solution,

Let the width of the sandbox be 'w'.

Since the sandbox is in shape of rectangle.

So we use the formula of area of rectangle.

Area = Length\times Width

On substituting the given values, we get;

\frac{1}{4}\times w=\frac{32}{3}\\

By cross multiplication method, we get;

w=\frac{32\times4}{3\times 1} =\frac{128}{3} =42\frac{2}{3} \ ft

Hence The width of the sandbox is 42\frac{2}{3} \ ft.

5 0
3 years ago
What are the odds of choosing a red marble from a bag that contains two blue marbles, one green marble and four red marbles?
Alisiya [41]

Answer:

4:3

Step-by-step explanation:

You count up all the red marbles which equals 4 and put them on one side, then you add up all the rest of the marbles same color or not which equals 3 and put it on the other side of the 4

4 0
3 years ago
A survey was done on the number of people in eachcar leaving a Shopping
ycow [4]

Answer:

a) Mode number = 2

b) Median  = 64.5  = 4 people in car as 0.5 in 64.5 is close to 76

              = 4 people

c) Mean number is  505/6 =   84.1666666667 then we see where 84.16 lies and see this lies with =  3 people

Step-by-step explanation:

45  is 1

198 is 2

121 is 3

76 is 4

52 is 5

13 is 6

Mode we look for the highest frequency and see 198 is the amount and 2 people in car is the subject, we give the subject and that is 2 people in car is the highest frequency and would be the mode.

Median is shown in the answer as we add up.

Mean we add up all frequency and divide by the amount of subjects = 6.

a) Mode number = 2

b) Median = 13, 45, 52, here 76, 121, 198    mid number between

             = (76-52 )/ 2  + (52)= 24 /2 + (52) = 12 + 52 + 0.5 if even number    

median  = 64.5  in a cumulative graph though which is not asked here it would be exactly half of 505 = 257.5 and interquartile we half again and add on 257.5 and show the range values interquartile as 1/2(252.5) and 1/2(252.5+ 505) = 126.25 as one value and 757.5 as the other by drawing a horizontal line from y axis to the points so vertical lines can represnt the cars median and interquartile.

c. Mean number is  505/6 =   84.1666666667 then we see where 84.16 lies and see this lies with 3 seats

For cumulative frequency we can relist numbers like this

45

45 +198 = 243

243 +  121 = 364

364+76 = 440

440 + 52 = 492

492+13 = 505 to plot graph.

But if there was group of measures or time etc then we would always plot the highest of each set for cumulative x axis and use the 2 values to see the ratio for histograms ie) 198/2 = 99 so that all values are read in equal proportions and 2 is a data below the graph not on the graph as 2 does not show as a box count just a name below on x axis. So we have to use the ratio as explained for histograms.

8 0
3 years ago
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