Answer:
-0.6
Step-by-step explanation:
-2.1--0.6= 1 1/2
2 negatives make a positive
Solution:
As, in a pack of cards, there are 26 black cards and 26 red cards.
Probability of an event = 
Probability of drawing a black card from a pack of 52 cards = 
As each card drawn is replaced,Each black card draw is an independent with another black card draw.
Probabilities of drawing a black card on each of four trial=
![\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=[\frac{1}{2}]^4](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%3D%5B%5Cfrac%7B1%7D%7B2%7D%5D%5E4)
Answer:
Step-by-step explanation:
Let <em>m</em> represent the amount of months that has passed.
Paulo started out with 20 coins.
Each month, he adds four more coins to his collection.
Therefore, the amount of coins added after <em>m</em> months will be given by 4<em>m</em>.
So, the amount of coins Paulo has total after <em>m</em> months can be given by:
We want to know when Paulo will have "more than" 50 coins in his collection. Therefore, we will use the "greater than" sign. Thus, our inequality will be:
Answer:
2000π in³/s
or 6283.2 in³/s
Step-by-step explanation:
dr/dt = 5
volume of sphere is given as
v = (4/3)(π)r³
differentitate wrt r
dv/dr = 3×4/3 ×π×r²
dv/dr = 4πr²
put r = 10
dv/dr = 4π(10)²
dv/dr = 400π
by chain rule
dv/dt = dv/dr ×dr/dt
dv/dt = 400π× 5
dv/dt = 2000π in³/s
or dv/dt =6283.2 in³/s
Complete Question:
A population proportion is 0.4. A sample of size 200 will be taken and the sample proportion p will be used to estimate the population proportion. Use z- table Round your answers to four decimal places. Do not round intermediate calculations. a. What is the probability that the sample proportion will be within ±0.03 of the population proportion? b. What is the probability that the sample proportion will be within ±0.08 of the population proportion?
Answer:
A) 0.61351
Step-by-step explanation:
Sample proportion = 0.4
Sample population = 200
A.) proprobaility that sample proportion 'p' is within ±0.03 of population proportion
Statistically:
P(0.4-0.03<p<0.4+0.03)
P[((0.4-0.03)-0.4)/√((0.4)(.6))/200 < z < ((0.4+0.03)-0.4)/√((0.4)(.6))/200
P[-0.03/0.0346410 < z < 0.03/0.0346410
P(−0.866025 < z < 0.866025)
P(z < - 0.8660) - P(z < 0.8660)
0.80675 - 0.19325
= 0.61351
B) proprobaility that sample proportion 'p' is within ±0.08 of population proportion
Statistically:
P(0.4-0.08<p<0.4+0.08)
P[((0.4-0.08)-0.4)/√((0.4)(.6))/200 < z < ((0.4+0.08)-0.4)/√((0.4)(.6))/200
P[-0.08/0.0346410 < z < 0.08/0.0346410
P(−2.3094 < z < 2.3094)
P(z < -2.3094 ) - P(z < 2.3094)
0.98954 - 0.010461
= 0.97908
