<span>Using the kinematic equations:
(final velocity)^2 = (initial velocity)^2 - 2 * acceleration * distance
Assuming the acceleration/deceleration on the car is constant from a constant force on the brakes. Converting from mph to m/s using 0.447 (so 34 mph is 15.2 m/s)
(0)^2 = (15.2)^2 - 2 * acceleration * 29
acceleration = 4.0 m/s^2
Had the car been going 105.4 mph (47.1 m/s)
(0)^2 = (47.1)^2 - 2 * 4 * distance
distance = 277 meters</span>
X/21 = y/33
xy = 21·33 = 693
The factors of 693 are 1,3,7,9,11,21,33,63,77,99,231,693.
(x,y) = (1,693), (693,1), (3,231), (231,3), (7,99), (99,7), (9,77), (77,9), (11,63), (63,11), (21,33), (33,21)
Answer:
Days- 9
Hours- 208
Minutes- 12,459
Step-by-step explanation:
For minutes,
747,533 ÷ 60 = 12,458.883 ≈ 12,459 minutes
For hours,
60 minutes to pass an hour so, 60 × 60 = 3600 minutes
747,533 ÷ 3600 = 207.648 ≈ 208 hours
For days,
1 day = 24 hours
1 hour = 3600 beats
24 × 3600 = 86400
747,533 ÷ 86400 = 8.652 ≈ 9 days
Answer:
the answer is c
Step-by-step explanation:
i just did it
Answer:
Step-by-step explanation:
Hello!
For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.
In this example the variable is:
X: height of a college student. (cm)
There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.
The option you have is to apply the Central Limit Theorem.
The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.
As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.
The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:
X[bar]~~N(μ;σ2/n)
Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:
98% CI
1 - α: 0.98
⇒α: 0.02
α/2: 0.01

X[bar] ± 
174.5 ± 
[172.22; 176.78]
With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].
I hope it helps!