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Gelneren [198K]
1 year ago
9

I have a limited time please help ;-;

Mathematics
2 answers:
RoseWind [281]1 year ago
7 0

Answer:$4

hope this helps



diamong [38]1 year ago
6 0
Jessica saves $4 more a week than Raphael. I took this quiz and it was correct.
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10 x 3 tens - unit form and standard form
vfiekz [6]
<span>10X3 tens in unit form is written:
10*3 tens = 30 tens = 300 units

10X3 tens in standard form:
10 x 3 tens = 10 x 30 = 300</span>
6 0
2 years ago
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At a construction site, a builder plans to cut down an 8-foot-by-9-foot piece of plywood. She plans to discard any pieces that s
AnnZ [28]

Answer:

12 i think

Step-by-step explanation:

6 0
2 years ago
The heights of the trees for sale at two nurseries are shown below. Heights of trees at Yard Works in feet : 7, 9, 7, 12, 5 Heig
Troyanec [42]

Answer:

The mean of tree heights at Yard Works is 8 feet and at Grow Station is 9 feet.The mean absolute deviation of the tree heights at both yards is 2.

Step-by-step explanation:

1). Height of the trees at Yard Works are = 7,9,7,12,5 feet

So mean height of the trees = (7+9+7+12+5)÷5

                                               = 40÷5 =8 feet

Standard deviation of the trees at Yard works = ∑(║(height of the tree-mean height of the tree))║/(number of trees)

(height of the tree-mean height of the tree)= ║(7-8)║+║(9-8)║+║(7-8)║+║(12-8)║+║(5-8)║ = (1)+1+(1)+4+(3)= 10

Therefore standard deviation = (10)/(5) =2

2). In the same way mean height of the trees at Grow Station=(9+11+6+12+7)/5= 45/5 = 9

Now we will calculate the mean deviation of the tress at Grow Station

= ∑║(height of the tree-mean height of the tree)║/(number of trees)

= ║(9-9)║+║(11-9)║+║(6-9)║+║(12-9)║+║(7-9)║/(5)

= (0+2+3+3+2)/5

= 10/5 =2

Therefore The mean of tree heights at Yard Works is 8 feet and at Grow Station is 9 feet.The mean absolute deviation of the tree heights at both yards is 2.

                                                             

7 0
2 years ago
Read 2 more answers
The figure below shows the dimensions of a city park in feet. Shannon thinks the area of city park is 73,084 square feet. Do you
Andreyy89

Answer:

We need to solve for the 4th side

4th side base = 75.5 -60.5 = 10 feet

4th side height = 16

4th side LENGTH^2 = 10^2 + 16^2

4th side = sq root (356) = 18.8679622641

Trapezoid Area = [(sum of the bases) / 2 ] * height

Trapezoid Area = [(136)/2] * 16

Trapezoid Area = 1,088 square feet, which is MUCH smaller

than 73,084

Step-by-step explanation:

3 0
1 year ago
Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases. (Assume that centra
Triss [41]

Answer:

Step-by-step explanation:

a) this involves 2 tails

The critical value is determined from the t distribution table.

α = 1 - 0.95 = 0.05

1 - α/2 = 1 - 0.05/2 = 1 - 0.025 = 0.975

Looking at 0.975 with df 10

The critical value is 2.228

b) α = 1 - 0.95 = 0.05

1 - α/2 = 1 - 0.05/2 = 1 - 0.025 = 0.975

Looking at 0.975 with df 20

The critical value is 2.086

c) α = 1 - 0.99 = 0.01

1 - α/2 = 1 - 0.01/2 = 1 - 0.005 = 0.995

Looking at 0.995 with df 20

The critical value is 2.845

d) α = 1 - 0.99 = 0.01

1 - α/2 = 1 - 0.01/2 = 1 - 0.005 = 0.995

Looking at 0.995 with df 60

The critical value is 2.660

e) 1 - α = 1 - 0.01 = 0.99

Looking at 0.99 with df 10

The critical value is 2.764

f) 1 - α = 1 - 0.025 = 0.975

Looking at 0.975 with df 5

The critical value is 2.571

6 0
1 year ago
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