In the first 1/6 hour of a rain storm, 1/10 inch of rain fell. If the rain continues to fall at the same rate, how much rain fell in 2 1/2 hours?

Answers

Answer 1

Answer:

Answer:

3/2 inches

Step-by-step explanation:

Related Questions

HELP ASAP PLSS! The graph represents the relationship between the amount of money a babysitter makes, m, and the hours the baby sitter works, h. Which function can be used to determine the amount of money the babysitter makes, d, after h hours? d=17h+5 d = 5h + 7 d = 7h +5 d=15h+7
PLEASE HELP!!!!Select the locations on the number line to plot the points 10/2 and -9/2
Please help answer this
If 150 mg of a drug is ordered IVP stat over 2 minutes, and the concentration of the drug is 75 mg/mL, find the total number of milliliters, and indicate how many milliliters you will administer per minute.A.1 mL total, 1 mL/min. B.2mL total, 1 mL/min.C.4 mL total, 2 mL/min.D.10 mL total, 5 mL/minute
Western Company allocates $10 overhead to products based on the number of machine hours used. The company uses a plantwide overhead rate with machine hours as the allocation base. Given the amounts below, how many machine hours does the company expect in department 2?

If the diameter of the circle is above 20mm what is the area

Answers

Answer:

The area would be 314.16 mm squared

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:

brainly.com/question/34083389

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6.2.Find the triangle bounded by the y-axis the line f(x)=9- 6/7x and the line perpendicular to it that passes through the origin

Answers

First.

We need to find the triangle bounded by the y-axis the line $f(x)=9-(6)/(7)x$ and the line perpendicular to it that passes through the origin.

This triangle is illustrated in the First Figure bellow. The triangle has been dotted with black dots. So, let's identify each part of this graph:

The y-axis has been identified by the green line.
The line $f(x)=9-(6)/(7)x$ has been identified by the red line.
To plot the line perpendicular to the previous one that passes through the origin, we need to know that the product of the slopes of two perpendicular lines is $m_(1)m_(2)=-1 \therefore Since \ m_(1)=-(6)/(7) \ then \ m_(2)=(7)/(6)$

Therefore, this line is the blue one and is written as:

$f(x)=(7)/(6)x$

Second.

a.

To find $y(t)$ we need to find the slope and y-intercept. So, taking two points:

$P_(1)(15,150) \ and \ P_(2)(25,450) \n \nm=(450-150)/(25-15)=30$

So:

$Using \ P_(1)(15,150) \n\ny=30t+b\n\n150=30(15)+b \therefore b=-300 \n\n\boxed{y(t)=30t-300}$

Third.

b.

The y-intercept is $b=-300$ and has been calculated above. This value represents the profit in thousands of dollars in 1980, that is, there is no any profit or there is a lost of $300.000.

Fourth.

c.

The x-intercept can be found when y=0 (year 1980), that is:

$y(t)=30t-300 \n\n0=30t-300 \therefore t=10$

This value shows that there is no any profit for the company at the year that represents t=10 (year 1990), because at this point the company starts earning money.

Fifth.

d.

The slope of this function is $m=30$

This value tells you that the profit increases $30.000 each year.

Hello,

Please, see the attached files.

Thanks.

Find the value of x that will make L||M

Answers

Answer:

x = 17

Step-by-step explanation:

When two lines are intersected by a transversal, the interior on the same side of the transversal are called co interior angles and they are supplementary.

6x + 8 + 4x + 2 = 180

6x + 4x + 8 + 2 = 180 {Combine like terms}

10x + 10 = 180 {Subtract 10 from both side}

10x = 180 - 10

10x = 170 {Divide both sides by 10}

x = 170/10

x = 17

What will the time be after 1 hour 5 minutes from 8:15 am

Answers

Answer:

9:20 am

Step-by-step explanation:

Add 1 hour

8:15 to 9:15

Add 5 minutes

9:15 to 9:20

Answer:

9:20 am

Step-by-step explanation:

So, lets go over two things.

Minutes and hours.

Minutes changes the second number.

You know how when a number goes from 9 to 10 how the ones place is set to 0, and the tens place goes up? Its the same with time, only when the number goes from 59 to 60, the hour goes up.

Hours changes the hours place, and when it hits 12, it resets to 1, and the words am go to pm, or pm goes to am.

In this case. we are moving the minutes place up by 5:

15+5=20

So the minutes place is 20, and does not change the hours place since it is below 60.

Next we have a increase in hours by 2:

8+1=9

So the hours place is 9, and does not reset or change the pm/am since its below 12.

Answer:

9:20am

Hope thias helps!

1. Simplify each of the following fractionsleaving each as an improper fraction, not a whole
number. Do not use your calculator!
4.50
5
3.2
8

Answers

Answers:

9/2

5/1

16/5

8/1

9/2
5/1
16/5
8/1
are the answers

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