Answer:
Step-by-step explanation:
h = -6
+20t + 4
I will use calculus, maybe that's not how you're supposed to do this
-12t +20 =0
12t = 20
t = 20 /12
t = 1 
t = 1 
there will be a max at 1.6666666666 seconds
-6*
+ 20 * 1.6666666666 +4
= 16.666666666666 + 33.333333333333 + 4
= - 16 + 33 
+ 4
= 20 feet max height ( not too high, for a rocket)
time of flight:
0 = -6 +20t + 4
use quadratic formula to find t
-20 +- sqrt [ 
- 4*(-6)*4 ] / 2*(-6)
-20 +- sqrt [400 + 96 ] / -12
-20 +- sqrt [496 ] / -12
-20 +- 22.27105 / -12
try the negative option 1st
-42.27105 / -12
3.522 seconds. time of flight
when will the rocket be at 12' ? :
12 = -6 +20t + 4
0 = -6 +20t -8
use quadratic formula again to find t
-20 +- sqrt [ - 4*(-6)*(-8) ] / 2*(-6)
-20 +- sqrt [ 400 - 192 ] / -12
-20 +- sqrt [208 ] / -12
-20 - 14.4222 / -12
-34.4222 / -12
2.8685 seconds ( on the way down)
and
-20 + 14.4222 / -12
-5.578 / - 12
0.4648 seconds ( on the way up )
Answer:
Step-by-step explanation:
Given that the observed frequencies for the outcomes as follows:
To check this we can use chi square goodness of fit test.
(Two tailed test at 5% significance level)
Assuming equally likely expected observations are found out and then chi square is calculated as (0-E)^2/E
Df = 6-1 =5
Outcome Frequency Expected frequency (Obs-exp)^2/Exp
1 36 34.83333333 0.03907496
2 30 34.83333333 0.670653907
3 41 34.83333333 1.091706539
4 40 34.83333333 0.766347687
5 23 34.83333333 4.019936204
6 39 34.83333333 0.498405104
209 209 7.086124402
p value =0.214
Since p >alpha, we accept null hypothesis
It appears that the loaded die does not behave differently than a fair die at 5% level of significance
Answer:
d
Step-by-step explanation:
from usatestprep:
The situation is not an example of uniform probability because freshmen, sophomores, juniors, and seniors do not have equal probabilities of being selected.; Uniform probability → equal probability of being selected
P(freshman) =
8/
26
; P(sophomore) =
7
/
26
; P(junior) =
6
/
26
; P(senior) =
5
/
26
; unequal probabilities → not uniform
58% is the answer. 7 dividing 12= 0.5833333
you much move the decimal over two places and you would get 58%
The probability that exactly six out of ten carry a smartphone when travelling is; <u><em>P(6) = 0.146</em></u>
The missing part of the question is;
The probability distribution of x is the binomial distribution with n = 10 and p = 0.75.
Calculate P(6)
- This is a binomial <em>probability distribution</em> problem that has a general formula as; P(x) = ⁿCₓ × pˣ × q⁽ⁿ ⁻ ˣ⁾
Where;
p is probability of success
q is probability of failure
n is number of experiments
In this question;
n = 10
p = 75% = 0.75
q = 1 - p
q = 1 - 0.75
q = 0.25
Since six passengers are randomly selected, then applying the formula earlier quoted, we have;
P(6) = ¹⁰C₆ × 0.75⁶ × 0.25⁽¹⁰ ⁻ ⁶⁾
P(6) = 0.146
Read more at; brainly.com/question/24239758
