Travis milks his cows each morning. He has never gotten fewer than 3 gallons of milk; however, he always gets fewer than 9 gallons of milk.

Answers

Answer 1

Answer: That’s the question?

Related Questions

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Gregory estimated the product of two decimals to be 24.Which could be the two decimals?
2.07(10.3)
4.02(6.8)
2.74(7.1)
5.9(3.92)

Answers

Answer:

it is the last one 5.9(3.92)

Step-by-step explanation:

i just took the exact same Assignment and got the question right soo....

D.) 5.9(3.92)

Explanation: (→ = estimated)

5.9 → 6

3.92 → 4

6 x 4 = 24

thus, D, 5.9(3.92), is your answer.

We know that if the probability of an event happening is 100%, then the event is a certainty. Can it be concluded that if there is a 50% chance of contracting a communicable disease through contact with an infected person, there would be a 100% chance of contracting the disease if 2 contacts were made with the infected person

Answers

Answer:

The correct answer to the following question will be "No". The further explanation is given below.

Step-by-step explanation:

Probability (Keeping the disease out of 1 contact)

= $0.5$

Probability (not keeping the disease out of 1 contact)

= $1-0.5$

= $0.5$

Now,

Probability (not keeping the disease out of 2 contact)

= Keeping the disease out of 1 contact × not keeping the disease out of 1 contact

On putting the estimated values, we get

= $0.5* 0.5$

= $0.25$

So that,

Probability (Keeping the disease out of 2 contact)

= $1-0.25$

= $0.75 \ i.e., 75 \ percent$

∴ Not 100%

Find the sum of the first 8 terms of the geometric sequence if the first term is 9 and the common ratio is -3.

Answers

hmmmm let's see $\bf \textit{sum of a finite geometric sequence}=S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\n\n\begin{cases}n=n^(th)\ term\na_1=\textit{first term}\nr=\textit{common ratio}\end{cases}\qquad thus\implies S_8=9\left( \cfrac{1-(-3)^8}{1-(-3)} \right)$

4272/6 use partial qutiants to divide

Answers

Answer:

712

Step-by-step explanation:

$4272/6= 4200/6+72/6\n4272/6= 700 +12\n4272/6=712$

Hope this helps :D

The rates of on-time flights for commercial jets are continuously tracked by the U.S. Department of Transportation. Recently, Southwest Air had the best reate with 80 % of its flights arriving on time. A test is conducted by randomly selecting 10 Southwest flights and observing whether they arrive on time. (a) Find the probability that at least 3 flights arrive late.

Answers

Answer:

There is a 32.22% probability that at least 3 flights arrive late.

Step-by-step explanation:

For each flight, there are only two possible outcomes. Either it arrives on time, or it arrives late. This means that we can solve this problem using binomial probability concepts.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_(n,x).\pi^(x).(1-\pi)^(n-x)$

In which $C_(n,x)$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_(n,x) = (n!)/(x!(n-x)!)$

And $\pi$ is the probability of X happening.

In this problem, we have that:

There are 10 flights, so $n = 10$ .

A success in this case is a flight being late. 80% of its flights arriving on time, so 100%-80% = 20% arrive late. This means that $\pi = 0.2$ .

(a) Find the probability that at least 3 flights arrive late.

Either less than 3 flights arrive late, or at least 3 arrive late. The sum of these probabilities is decimal 1. This means that:

$P(X < 3) + P(X \geq 3) = 1$

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_(n,x).\pi^(x).(1-\pi)^(n-x)$

$P(X = 0) = C_(10,0).(0.2)^(0).(0.8)^(10) = 0.1074$

$P(X = 1) = C_(10,1).(0.2)^(1).(0.8)^(9) = 0.2684$

$P(X = 2) = C_(10,2).(0.2)^(2).(0.8)^(8) = 0.3020$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1074 + 0.2684 + 0.3020 = 0.6778$

Finally

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.6778 = 0.3222$

There is a 32.22% probability that at least 3 flights arrive late.

Final answer:

The problem is solved by calculating the probability of the complementary event (0,1,2 flights arriving late) using the binomial distribution, then subtracting this from 1 to find the probability of at least 3 flights arriving late.

Explanation:

This problem is typically solved by using a binomial probability formula, which is used when there are exactly two mutually exclusive outcomes of a trial, often referred to as 'success' and 'failure'.
Here, our 'success' is a flight arriving late. The probability of success, denoted as p, is thus 20% or 0.2 (since 80% arrive on time, then 100%-80% = 20% arrive late). The number of trials, denoted as n, is 10 (the number of randomly selected flights).
We want to find the probability that at least 3 flights arrive late, in other words, 3,4,...,10 flights arrive late. The problem can be solved easier by considering the complementary event: 0,1,2 flights arrive late. Then subtract the sum of these probabilities from 1.

The binomial probability of exactly k successes in n trials is given by:

P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
Where C(n, k) is the binomial coefficient, meaning choosing k successes from n trials.
We calculate like so:
P(X=0) = C(10, 0) * (0.2)^0 * (0.8)^10
P(X=1) = C(10, 1) * (0.2)^1 * (0.8)^9
P(X=2) = C(10, 2) * (0.2)^2 * (0.8)^8
Sum these up and subtract from 1 to get the probability that at least 3 flights arrive late. This gives the solution to the question.

Learn more about binomial probability here:

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Assume f(x) = ax^2+ bx + c Which one is true? f is a parabola that opens sideways to the right. f is a parabola that opens sideways to the left. f is a parabola that opens downward. f is a parabola that opens upward.

Answers

Step-by-step explanation:

f is a parabola that opens upward.

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