Answer:
Pyramid=5
Cone=1
Cylinder=3
Triangular Prism=4
Rectangular Prism=2
Step-by-step explanation:
One part of six parts is red, so just make it a fraction like so: 1/6 because 1 side is red out of six total parts.
Answer:
The area after 9 years will be 1,234 km^2
Step-by-step explanation:
In this question, we are tasked with calculating what the area of a certain forest that decreases at a certain percentage would be after some years.
To answer this question, we shall be using an exponential approximation.
Now, to use this exponential approximation, we shall be needing a supporting exponential mathematical equation.
This can be written as;
A = I(1-r)^t
where A is the new area we are looking for
I is the initial area which is 1700 according to the question
r is the rate of decrease which is 3.5% = 3.5/100 = 0.035
t is time which is 9 years according to the question
We plug these values and have the following;
A = 1700(1-0.035)^9
A = 1700(0.965)^9
A = 1,233.66
This is 1,234 km^2 to the nearest square kilometer
Answer:
11,4 > 11 > 9,8 > 1
Step-by-step explanation:
To list the numbers from smallest to largest we need to determine the value of them in same/similar format
11 and 1 is in same format
The approximate value of 
9.8
the root of 
= 11
the root of 
is = 11.4 approximately
Now we have them as real numbers we can evaluate
11,4 > 11 > 9,8 > 1 we don't need to write 11 twice but if needed we can write is as:
11,4 > 11 = 11 > 9,8 > 1
Best Answer:<span> </span><span>A. Mode
B. Median
C. Mean
the other two are not measures of central tendency.
Data:
4+9/12, 5+3/12, 5+4/12,5+7/12,5+3/12, 5+9/12,6+0/12, 5+4/12,6+0/12, 6+1/12,5+7/12, 5+5/12, 4+5/12
4.4167, 4.75, 5.25{2}, 5.3333{2}, 5.4167, 5.5833{2}, 5.75, 6{2}, 6.0833
n= 13; Σx= 70.75; Σx²= 387.8403; Σx³= 2140.408; Σx^4= 11886.2257
σ= 0.4638; σn-1= 0.4828; Variance=σ²= 0.2152; (σn-1)²= 0.2331
Mean(μ)= 5.4423; range= 1.6667; mid-range= 5.25; median= 5.4167; No Mode Probably right skewed.
Five number summary: Q0= 4.4167, Q1= 5.25, Q2= 5.4167, Q3= 5.875, Q4= 6.0833, IQR= 4.875</span>
