Answer:
she starts with $25
every three games she spends $1.50 (.50 per game)
Step-by-step explanation:
i mean i cant really explain it, its kinda self explanatory
Answer:
1. 18
2. 2412
Step-by-step explanation:
1. Monday 3
Tuesday 6
Wensday 9
Thursday 12
Friday 15
Sunday 18
_____________________________________________________
2.
603*4=2412
Step-by-step explanation:
The solution to this problem is very much similar to your previous ones, already answered by Sqdancefan.
Given:
mean, mu = 3550 lbs (hope I read the first five correctly, and it's not a six)
standard deviation, sigma = 870 lbs
weights are normally distributed, and assume large samples.
Probability to be estimated between W1=2800 and W2=4500 lbs.
Solution:
We calculate Z-scores for each of the limits in order to estimate probabilities from tables.
For W1 (lower limit),
Z1=(W1-mu)/sigma = (2800 - 3550)/870 = -.862069
From tables, P(Z<Z1) = 0.194325
For W2 (upper limit):
Z2=(W2-mu)/sigma = (4500-3550)/879 = 1.091954
From tables, P(Z<Z2) = 0.862573
Therefore probability that weight is between W1 and W2 is
P( W1 < W < W2 )
= P(Z1 < Z < Z2)
= P(Z<Z2) - P(Z<Z1)
= 0.862573 - 0.194325
= 0.668248
= 0.67 (to the hundredth)
Answer: Speed = 5.6 mph
Oxygen consumption = 71.6 mL/lb/min
Step-by-step explanation: For the oxygen consumption to be the same, functions must be equal:
f(x) = g(x)

Resolving:





x=0
5x - 28 = 0

x = 5.6
<u>The speed when the oxygen consuption is the same is </u><u>5.6 mph</u><u>.</u>
For the level of oxygen consumption:
f(5.6) = g(5.6)
g(5.6) = 11*5.6 + 10
g(5.6) = 71.6
<u>The level of oxygen consumption is </u><u>71.6 mL/lb/min</u>