Answer:
1 4/5 or 1.8 pints are white
Step-by-step explanation:
1/4 of 7 1/5 = 1/4 x 36/5 = 36/20 or 9/5, which is 1 4/5
275 beacues if u and 50 five times it would of been 250 but u an the five two the 50 and u and 55 five times and u get 275
Answer:
The shadow is decreasing at the rate of 3.55 inch/min
Step-by-step explanation:
The height of the building = 60ft
The shadow of the building on the level ground is 25ft long
Ѳ is increasing at the rate of 0.24°/min
Using SOHCAHTOA,
Tan Ѳ = opposite/ adjacent
= height of the building / length of the shadow
Tan Ѳ = h/x
X= h/tan Ѳ
Recall that tan Ѳ = sin Ѳ/cos Ѳ
X= h/x (sin Ѳ/cos Ѳ)
Differentiate with respect to t
dx/dt = (-h/sin²Ѳ)dѲ/dt
When x= 25ft
tanѲ = h/x
= 60/25
Ѳ= tan^-1(60/25)
= 67.38°
dѲ/dt= 0.24°/min
Convert the height in ft to inches
1 ft = 12 inches
Therefore, 60ft = 60*12
= 720 inches
Convert degree/min to radian/min
1°= 0.0175radian
Therefore, 0.24° = 0.24 * 0.0175
= 0.0042 radian/min
Recall that
dx/dt = (-h/sin²Ѳ)dѲ/dt
= (-720/sin²(67.38))*0.0042
= (-720/0.8521)*0.0042
-3.55 inch/min
Therefore, the rate at which the length of the shadow of the building decreases is 3.55 inches/min
It is:
40+1+0.3+0.02=41.32
Answer:
The sample space would be infinite
{HH, TTTHH, TTTTHH, TTTTTTTHH.......................}
1/8
Step-by-step explanation:
The sample space would be infinite
{HH, TTTHH, TTTTHH, TTTTTTTHH.......................}.
This is because the number of times the coin would be flipped is not specified. So, you can keep flipping the coin forever.
If the coin is tossed four times then the sample space would be
{HHHH HTHH THHH HTHT
HHHT HTTH TTHH THTH
HHTT HHTH TTTH THHT
HTTT TTTT TTHT THTT}
HTHH and TTHH are the only two cases where two consecutive tosses will result in two heads
Probability that the coin will be tossed four times is 2/16 = 1/8