Answer: X=6.
Step-by-step explanation:
16x-8+2x=100
18x-8=100
18x=100+8
18x=108
x=108/18
x=6
Answer:
A: No, because the bike order does not meet the restrictions of 4c + 6a ≤ 120 and 4c + 4a ≤ 100
Step-by-step explanation:
c = child bike
a = adult bike
Child bike requirement:
4 hours build time
4 hours test time
Adult bike requirement:
6 hours build time
4 hours test time
Available Building time per week ≤ 120 hours
Available Testing time per week ≤ 100 hours
Therefore, company's work schedule is;
Building:
4c + 6a ≤ 120
Testing:
4c + 4a ≤ 100
Combined:
4c + 6a ≤ 120 and 4c + 4a ≤ 100
Building 20 Child bike and 6 adult bike in a week:
4(20) + 6(6) ≤ 120 and 4(20) + 4(6) ≤ 100
80 + 36 ≤ 120 - - - - 80 + 24 > 100
Therefore Building 20 Child bike and 6 adult bike in a week is not feasible.
probability that a randomly chosen college student either has a job or lives on campus.
What is probability?
- Probability is an area of mathematics that deals with numerical descriptions of how probable an event is to occur or how likely a statement is to be true.
- The probability of an event is a number between 0 and 1, where 0 denotes the event's impossibility and 1 represents certainty.
To find the probability that a randomly chosen college student either has a job or lives on campus:
Given: 85% of college students have a job, 54% live on campus, and 42% have a job and live on campus.
So, out of 100 students, 42 students either has a job or live on campus.
43 college students have a job and 12 live on campus.
So, 43 + 42 + 12 = 85 + 54 - 42
85 + 12 = 139 - 42
97 = 97
Therefore, probability that a randomly chosen college student either has a job or lives on campus.
Know more about probability here:
brainly.com/question/24756209
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X axis because of its sequence that it is involved with a reflection which is a transformation
Answer:
Step-by-step explanation:
Given that,
- p ( probability that the child has disease) = 25% = 0.25
- n = number of children = 3
The probability mass function of binomial distribution is,
- (P = X) = (nCx) X (p)^x X (1 - p)^n-x ; x = 0, 1, 2 ,3
- = 3Cx X (0.25)^x X (1 - 0.25)^3-x ; ( n = 3, p = 0.25
a) P ( two will have disease)
p ( X = 2) = 3C2 X (0.250^2 X (1 - 0.25) ^3-2
= 0.1406
b) P ( none will have disease)
p (X = 0) = 3C0 X (0.25)^0 X (1 - 0.25)^3-0
= 0.4219
c) P (neither having the disease nor being a carrier) = 25% = 0.25
The probability that at least one will neither having the disease nor being a carrier ;
P(X> or equals to) = 1 - P(X < 1)
= 1 - P( X = 0)
= 1 - 3C0 X (0.25)^0 X (1 - 0.25)^3-0
= 0.5781
d) p( the first child with the disease will the be 3rd child)
P(X = x) = (1-p)^x-1 X p
p( X= x) = ( 1 - 0.25 )^x -1 X 0.25
for third child = P(X = 3) = (1 - 0.25)^3-1 X (0.25)
= 0.1406