The answer would be (-5+_sqroot(-3))/2 which are answers A and F
Answer:
The rate is 1 ÷ 2 yards per hour
Step-by-step explanation:
The computation of the rate does the amount of yarn changes is given below:
= 3 3 ÷ 4 - 1 1 ÷8
= 15 ÷ 4 - 9 ÷ 8
= 21 ÷ 8 yards
Now divide the above fraction by the number of hours
= 21 ÷ 8 yards ÷ 5 1 ÷ 4
= 21 ÷ 8 × 4 ÷ 21
= 1 ÷ 2 yards per hour
Hence, the rate is 1 ÷ 2 yards per hour
The same would be relevant
Answer:
B. Yes, there are at least 10 people with weak bones and 10 people with strong bones in each group.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean
and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation 
The correct answer is:
B. Yes, there are at least 10 people with weak bones and 10 people with strong bones in each group.
Are there answers you can provide?
You read the number "sixty-eight and 18 thousandths" \
Hope that helps!
Answer:
a.) f(x) =
where 90 < x < 120
b.) 
c.) 
d.) 
Step-by-step explanation:
Let
X be a uniform random variable that denotes the actual charging time of battery.
Given that, the actual recharging time required is uniformly distributed between 90 and 120 minutes.
⇒X ≈ ∪ ( 90, 120 )
a.)
Probability density function , f (x) =
where 90 < x < 120
b.)
P(x < 110) = 
= ![\frac{1}{30}[x]\limits^{110}_{90} = \frac{1}{30} [ 110 - 90 ] = \frac{1}{30} [ 20] = \frac{2}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B30%7D%5Bx%5D%5Climits%5E%7B110%7D_%7B90%7D%20%20%3D%20%5Cfrac%7B1%7D%7B30%7D%20%5B%20110%20-%2090%20%5D%20%3D%20%5Cfrac%7B1%7D%7B30%7D%20%5B%2020%5D%20%3D%20%5Cfrac%7B2%7D%7B3%7D)
c.)
P(x > 100 ) = 
= ![\frac{1}{30}[x]\limits^{120}_{100} = \frac{1}{30} [ 120 - 100 ] = \frac{1}{30} [ 20] = \frac{2}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B30%7D%5Bx%5D%5Climits%5E%7B120%7D_%7B100%7D%20%20%3D%20%5Cfrac%7B1%7D%7B30%7D%20%5B%20120%20-%20100%20%5D%20%3D%20%5Cfrac%7B1%7D%7B30%7D%20%5B%2020%5D%20%3D%20%5Cfrac%7B2%7D%7B3%7D)
d.)
P(95 < x< 110) = 
= ![\frac{1}{30}[x]\limits^{110}_{95} = \frac{1}{30} [ 110 - 95 ] = \frac{1}{30} [ 15] = \frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B30%7D%5Bx%5D%5Climits%5E%7B110%7D_%7B95%7D%20%20%3D%20%5Cfrac%7B1%7D%7B30%7D%20%5B%20110%20-%2095%20%5D%20%3D%20%5Cfrac%7B1%7D%7B30%7D%20%5B%2015%5D%20%3D%20%5Cfrac%7B1%7D%7B2%7D)