Complete question :
The Venn diagram relating to the question can be found in the picture attached below :
Answer:
A.) 15 ; b.) 17 ; c.) 20 ; d.) 19 ; e.) 55 ; 67; 76 ;100 ; F.) 369
Step-by-step explanation:
Let :
Cars = C ; Motorcycle = M ; Tricycle = T ; Walking = W
a) How many students ride in a tricycle, motorcycle and car going to their school
Intersection of the 3 modes;
(C n M n T) = 15 ; it is the number which sits in between all the three circles.
B.) How many students ride in both a motorcycle and a tricycle?
(M n T) = 17 ; number in the middle of both circles representing motorcycle and tricycle
C.) How many students ride in both a motorcycle and a car?
(M n C) = 20 ; number in the middle of both circles representing motorcycle and Car
D) How many students ride in both a car and tricycle?
(C n T) = 19 ; number in the middle of both circles representing Car and tricycle
e.)How many students go to school
in a car only = 55
in a motorcycle only = 67
Tricycle only = 76
Walking = 100
F.) How many Grade Seven students of Koronadal National Comprehensive High School are there in all?
(100 + 67 + 76 + 55 + 19 + 20 + 17 + 15) = 369
Answer:
The expected frequency for the cell E2,2 is = 33
Step-by-step explanation:
The given data is
Gender | 0-30| 30-60| 60-90| 90+ Totals
Male | 23 | 35 | 76 | 46 180
<u> Female | 31 | 42 | 46 | 16 135 </u>
<u>Totals 54 77 122 62 315 </u>
<u />
The expected frequency for the cell E2,2 is :
Expected for the (30-60 box for females) = Total of (30-6)/ (total )* females
= ( 77/315)135= 33
Here p= 77/315 and n= 135
therefor X= pn = 33
χ²= (33-42)²/33= 2.455 ( for the single value of E2,2=33)
Expected for the (30-60 box for males) = Total of (30-6)/ (total )* males
= ( 77/315)180= 44
χ²= (44-42)²/44= 0.09
The critical region is χ² (0.05) 3 ≥ χ²= 11.34
Let the null and alternate hypothesis be
H0:the number of minutes spent online per day is not related to gender
against
Ha: the number of minutes spent online per day is related to gender
The single value of χ² for E2,2 = 2.45 is less than the critical value of 11.34.
Answer:
Change in area=24
-48
Step-by-step explanation:
Let s will be the side of square and r will be the radius of circle.
Then two given conditions are
1)dr/dt=2 m/s
2)ds/dt=1 m/s
Area enclosed=(Area of square)-(Area of circle)
Area of square=
Area of circle=
Area enclosed=
dA/dt=2r(dr/dt)-2s(ds/dt)
At s=24,and r=6
dA/dt=2()(6)(2)-2(24)(1)
Change in area=24-48
I’m pretty sure This rounds to 900
I think is inside and then outside. Not sure
Hope this helps!
