9 + 1.34 + .5 (3.50 +1.74) How would you simplify the expression? Explain your steps.
Answers
(3.50+1.74)= 5.24
Now the equation is:
9+1.34+.5(5.24)
2. Multiply .5(5.24)
9+1.34+2.62=
12.96
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6 ft.
6
ft.
Answers
Answer:
3 ft
Step-by-step explanation:
Answers
Answer:
-8 and 6
Step-by-step explanation:
Coordinate of point M = -1
Distance between M and a point N = MN = 7
Therefore, the possible coordinates of N can either be:
To the left => -1 - 7 = -8
Or
To the right => -1 + 7 = 6
Check:
If M = -1, and N = -8, MN = |-1 -(-8)| = |-1 + 8| = 7
Or
If M = -1, and N = 6, MN = |-1 - 6| = |-7| = 7
So, our answer is right.
Possible coordinates of point N are -8 and 6
Answers
( x - 7 ) + 2........................
Answers
She can make 2 goodie bags.
13 lollipops
17 tootsie rolls
The GCF (greatest common factor) of 26 and 14 is 2; therefore making the greatest number of goodie bags she can make is 2.
If you divide both the numbers (26,14) by 2, you get 13 and 17.
Please mark branliest if this helped
Answers
Let n represent the unknown number.
14.4 % x n= 10.4
n = 10.4 / 14.4%
n = 72.22222
n = ~72
Hope this helps!
Answers
Analyzing a sample of 14 flights at Denver International Airport, the probability of 10 or more flights arriving on time is 0.3783, and the probability of 11 or more flights arriving on time is 0.2142, which is not considered unusual.
(a) All 12 of the flights were on time.
(b) Exactly 10 of the flights were on time.
(c) 10 or more of the flights were on time.
(d) Would it be unusual for 11 or more of the flights to be on time?
We can use the binomial probability formula to solve this problem. The binomial probability formula is:
P(k successes in n trials) = (n choose k) * *
where:
n is the number of trials
k is the number of successes
p is the probability of success
q is the probability of failure
In this case, n = 14, p = 0.85, and q = 0.15.
(a) To find the probability that all 12 of the flights were on time, we can plug k = 12 into the binomial probability formula:
P(12 successes in 14 trials) = (14 choose 12) * *
Using a calculator, we can find that this probability is approximately 0.0032.
(b) To find the probability that exactly 10 of the flights were on time, we can plug k = 10 into the binomial probability formula:
P(10 successes in 14 trials) = (14 choose 10) * *
Using a calculator, we can find that this probability is approximately 0.1022.
(c) To find the probability that 10 or more of the flights were on time, we can add up the probabilities of 10, 11, 12, 13, and 14 successes:
P(10 or more successes) = P(10 successes) + P(11 successes) + P(12 successes) + P(13 successes) + P(14 successes)
Using a calculator, we can find that this probability is approximately 0.3783.
(d) To determine whether it would be unusual for 11 or more of the flights to be on time, we can find the probability of this event and compare it to a common threshold for unusualness, such as 0.05.
P(11 or more successes) = P(11 successes) + P(12 successes) + P(13 successes) + P(14 successes)
Using a calculator, we can find that this probability is approximately 0.2142. This probability is greater than 0.05, so it would not be considered unusual for 11 or more of the flights to be on time.
Final answer:
This problem can be approached as a binomial distribution. The probability of a particular number of flights on time is calculated using the binomial probability formula. Determining 'unusual' can be subjective but normally a probability less than 0.05 is considered unusual.
Explanation:
This problem is a binomial probability problem because we have a binary circumstance (flight is either on time or it isn't) and a fixed number of trials (14 flights). The binomial probability formula is P(X=k) = C(n, k) * (p^k) * ((1 - p)^(n - k)) where n is the number of trials, k is the number of successful trials, p is the probability of success on a single trial, and C(n, k) represents the number of combinations of n items taken k at a time.
(a) For all 12 flights on time, it seems there's a typo; there are 14 flights in the sample. We can't calculate for 12 out of 14 flights without the rest of the information.
(b) For exactly 10 flights, we use n=14, k=10, p=0.85: P(X=10) = C(14, 10) * (0.85^10) * ((1 - 0.85)^(14 - 10)).
(c) For 10 or more flights on time, it's the sum of the probabilities for 10, 11, 12, 13, and 14 flights on time.
(d) For determining whether 11 or more on-time flights is unusual, it depends on the specific context, but we could consider it unusual if the probability is less than 0.05.
Learn more about Binomial Probability here:
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