Classify the triangle below explain how you know (20points)
1 answer:
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Answer:
About 296 households have more than three children.
Step-by-step explanation:
The complete question is:
Carmen used a random number generator to simulate a survey of how many children live in the households in her town. There are 1,346 unique addresses in her town with numbers ranging from zero to five children. The results of 50 randomly generated households are shown below. Children in 50 Households Number of Children Number of Households 0 5 1 11 2 13 3 10 4 9 5 2 Using a proportion, what can you infer about the number of households in her town that have more than three children? About 242 households have more than three children. About 269 households have more than three children. About 296 households have more than three children. About 565 households have more than three children.
Solution:
The data provided for the number of children and number of households is:
Number of Children (<em>X</em>) Number of Households (<em>f</em> (X))
0 5
1 11
2 13
3 10
4 9
5 2
TOTAL 50
Compute the probability of households having more than three children as follows:
P (More than 3 children) = P (4 children) + P (5 children)
Let the random variable <em>Y</em> be defined as the number of households having more than three children.
The probability of the random variable <em>Y </em>is, <em>p</em> = 0.22.
The entire population, i.e. total number of addresses in Carmen's town with numbers ranging from zero to five children, consists of <em>N</em> = 1,346 households.
Compute the expected number of households having more than three children as follows:
Thus, about 296 households have more than three children.
Answer:
2598 square kilometers
Step-by-step explanation:
Hello
Step 1
year one
using a rule of three is possible to find how much is 8.75 od 4500 km2
Let
if
4500 km2 ⇒ 100$
x?km2 ⇒8.75
do the relation and isolate x
at the end of the year one, the area will be
4500-393.75=4106.25
this will be the initial area for the year 2.
Step 2
repite the step 1 with area initial =4106.25 km2
4106.25 km2 ⇒ 100$
x?km2 ⇒8.75
do the relation and isolate x
at the end of the year 2, the area will be
4106-359.29=3746.70
this will be the initial area for the year 3.
Step 3
repite the step 1 with area initial =4106.25 km2
3746.70 km2 ⇒ 100$
x?km2 ⇒8.75
do the relation and isolate x
at the end of the year 3, the area will be
3746.70-327.83=3419.09
this will be the initial area for the year 4.
Step 4
year four
repite the step 1 with area initial =3419.09 km2
3419.09 km2 ⇒ 100$
x?km2 ⇒8.75
do the relation and isolate x
at the end of the year 4, the area will be
3419.09-299.173=3119.82
this will be the initial area for the year 5.
Step 5
year five
repite the step 1 with area initial =3119.82 km2
3119.82 km2 ⇒ 100$
x?km2 ⇒8.75
do the relation and isolate x
at the end of the year 5, the area will be
3119.82-272.99=2846.92
this will be the initial area for the year 6.
Step 6
year six
repite the step 1 with area initial =2846.92km2
2846.92 km2 ⇒ 100$
x?km2 ⇒8.75
do the relation and isolate x
at the end of the year six, the area will be
2846.92-249.10=2597.82 square kilometers
Have a great day.