10. Sarah has a new job of slicing and
stocking the lunchmeat in a deli case.
She keeps track of how many pounds
she puts in the case, while her manager
keeps track of the number of pounds
they sell from the case.
Sarah's List
6
10.60
Day
143.46 1
2
3
bropez
10
bro
Day
Number of Pounds
Added to Case
0.5
10
0.5
10de
2
erit
aby
List ole qu zmlwz
Manager's List
Number of Pounds Soldot
exom S
from Case
wan 11 svib ed EdW
24/4
3
If there were 12 pounds in the case
when Sarah started on Day 1, how
many pounds will be in the case at the
start of Day 4?

Answer:

Kai is correct and Leila is incorrect. The movie is actually 10,800 seconds long.

Step-by-step explanation:

1 hour = 60 minutes

1 minute = 60 seconds

Therefore, 1 hour = 60 x 60 = 3600 seconds

So, the movie's length in seconds is:

3 hours x 3600 seconds/hour = 10,800 seconds

Leila says the movie is less than 10,000 seconds, which is not correct, since the movie is actually longer than 10,000 seconds.

Kai says the movie is more than 10,000 seconds, which is correct.

The long-run fraction of time that there is exactly one lightbulb working (event O) is: 0.228.

Let's use the following notation:

Let W denote the event that both lightbulbs are working,

let O denote the event that one lightbulb is working, and

let B denote the event that both lightbulbs are burnt out.

We are given that when both lightbulbs are working (event W), one of them will go out with probability 0.03.

Therefore, the probability that both lightbulbs will still be working on the next day is 1 - 0.03 = 0.97.

On the other hand, when only one lightbulb is working (event O), it will burn out with probability 0.07, and the other lightbulb is already burnt out.

Hence, the probability of moving from O to B is 1.

We can set up the following system of equations to model the probabilities of being in each state on the next day:

P(W) = 0.97P(W) + 0.5P(O)

P(O) = 0.03P(W) + 0.93P(O) + 1P(B)

P(B) = 0.07P(O)

Note that in the first equation, we use 0.97 because the probability of staying in W is 0.97, and the probability of moving to O is 0.5 (because there are two ways for one of the lightbulbs to go out).

Simplifying the system of equations, we get:

0.03P(W) - 0.5P(O) = 0

-0.03P(W) + 0.07P(O) - 1P(B) = 0

0P(W) - 0.07P(O) + 1P(B) = 0

Solving for P(O), we get:

P(O) = 0.3P(W)

Substituting this into the second equation, we get:

-0.03P(W) + 0.07(0.3P(W)) - P(B) = 0

Simplifying, we get:

P(B) = 0.004P(W)

We also know that the sum of the probabilities of being in each state must be 1:

P(W) + P(O) + P(B) = 1

Substituting the expressions for P(O) and P(B), we get:

P(W) + 0.3P(W) + 0.004P(W) = 1

Solving for P(W), we get:

P(W) = 0.762

Therefore, the long-run fraction of time that there is exactly one lightbulb working (event O) is:

P(O) = 0.3P(W) = 0.228.

Approximately 22.8% of the time, there will be exactly one lightbulb working.

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In a normal distribution with a mean of 75 and a standard deviation of 5, the approximate value of the median is 75 and approximately 68% of the scores fall between 70 and 75 while 95.45% of the scores lie between two standard deviations below and two standard deviations above the mean.

Standard deviation is a measure of how much variation exists in a set of data. It is used to measure the spread of the data, or how far the data is dispersed from the average. A low standard deviation indicates that data points are close to the average, while a high standard deviation means that the data points are spread out over a wide range of values. Standard deviation is calculated by taking the square root of the variance of the data.

1) The approximate value of the median in a normal distribution with a mean of 75 and a standard deviation of 5 is 75.

2) Approximately 68% of the scores fall between 70 and 75. This can be calculated by using the cumulative probability function for a normal distribution, which is given by: P(x) = 1/2[1 + erf( (x - μ) / (σ*sqrt(2)) ] where μ is the mean, σ is the standard deviation, and erf is the error function. In this case, the cumulative probability of 70 is 0.5 and the cumulative probability of 75 is 0.8413, so the difference of 0.3413 gives the approximate percentage of scores between 70 and 75.

3) Approximately 95.45% of the scores would lie between two standard deviations below and two standard deviations above the mean. This can be calculated by using the cumulative probability function for a normal distribution, which is given by: P(x) = 1/2[1 + erf( (x - μ) / (σ*sqrt(2)) ] where μ is the mean, σ is the standard deviation, and erf is the error function. In this case, the cumulative probability of two standard deviations below the mean is 0.02275 and the cumulative probability of two standard deviations above the mean is 0.97725, so the difference of 0.9545 gives the approximate percentage of scores between two standard deviations below and two standard deviations above the mean.

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Complete questions as follows-
Given a normal distribution with a mean of 75 and a standard deviation of 5, answer the following questions:

1) What is the approximate value of the median?

2) What percentage of scores fall between 70 and 75?

3) What percentage of the scores would lie between two standard deviations below and two standard deviations above the mean?

Answer:

Step-by-step explanation:

If the solutions are x = -3 and x = -1, then (x - 3) (x - 1) will give us our answer. Using the FOIL method,

(x - 3) (x - 1)

x^2 - x - 3x + 3

x^3 - 4x + 3 = 0

Your answer is 3

Emma has a credit card with 15% annual interest compounded monthly. The exponential expression is 300(1+0.15/12)^(12*t). She owes $509 in 4 years. It's about $464. Time it will take for her balance to reach $450, which is about 5.6 years. The expression to evaluate her balance with an annual compounding interest rate is 300(1+0.15)^t = 300(1.15)^t. If the card were compounded annually, she would owe about $459 in 4 years and it would take about 6.5 years for her balance to reach $450.

P = $300 (initial balance)

r = 0.15 (annual interest rate)

n = 12 (monthly compounding)

The simplified exponential expression is

300(1+0.15/12)^(12*t)

Using the formula from Part A with t = 4

300(1+0.15/12)^(12*4) ≈ $509

Emma will owe approximately $509 in 4 years if she doesn't make any payments toward the balance.

The equation for the situation is

300(1+0.15/12)^(12*t) = 450

Taking the logarithm of both sides and solving for t:

t = log(450/300) / (12*log(1+0.15/12)) ≈ 5.6 years

It will take about 5.6 years for Emma's balance to reach $450 assuming she doesn't make any payments toward it.

The expression Emma could use to evaluate her balance with an annual compounding interest rate is

300(1+0.15)^t = 300(1.15)^t

After 4 years, Emma would owe approximately $459 for her original purchase of $300.

It would take around 6.5 years for her balance to increase from $300 to $450.

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If after putting in 2/3 of total-drops needed, Alberto stop and answer his phone, then the number of drops that Alberto still need to add to water is 20 drops.

The directions say to use "1 drop" for each 1/4 gallon of water, and Alberto has 15 gallons of water in the tank,

So, the "total-number" of drops needed is :

⇒ Total number of drops needed = (1 drop per 1/4 gallon) × (15 gallons) × (4 quarters per gallon),

⇒ Total number of "drops-needed" is = 60 drops,

Next, we calculate "2/3" of "total-drops" needed,

To find out how many drops Alberto has already added,

We calculate 2/3 of the "total-drops" needed,

⇒ 2/3 of total drops = (2/3) × (total number of drops needed),

⇒ 2/3 of 60 drops = (2/3) × 60 = 40 drops,

So, Alberto has already added 40 drops of water-cleaning drops,

To calculate how many drops Alberto still needs to add, we subtract drops he already added from total drops needed,

⇒ Drops still needed = (Total drops needed) - (Drops already added),

⇒ 60 drops - 40 drops = 20 drops

Therefore, Alberto still needs to add 20 drops.

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Hence correct option or expression are D and F.

its branches of mathematics. The  arithmetic deals with numbers and mathematical procedures. Math think how to add, subtract, multiply, and divide two or more  numbers. Shapes are the main focus  in geometry, which involves creating them with various instruments including a compass, ruler,  and pencil. Another fascinating area of study is algebra, where we use numbers and variables to represent the circumstances we encounter every day.

A mathematical function called an exponential function is employed frequently in everyday life. It is mostly used to compute investments, model populations, determine exponential decline or exponential growth,  and so forth. You will discover  the formulas, guidelines, characteristics, graphs, derivatives, exponential series, and examples of exponential functions in this article.

use,

[tex]\frac{a^{m} }{a^{n} } =a^{m-n}[/tex]

so,

[tex]\frac{b^{-2} }{b^{-6} } =b^{-2+6}\\=b^{4} or \frac{1}{b^{-4} }[/tex]

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Answer: To check if v=10 is a solution to the equation 0.2v = 1.2, we can substitute v=10 into the equation and see if the equation holds true:

0.2v = 1.2

0.2(10) = 1.2

2 = 1.2

This is not true, since 2 is not equal to 1.2. Therefore, v=10 is not a solution to the equation 0.2v = 1.2.

Step-by-step explanation:

Answer:

solution

Step-by-step explanation:

This means that if the average weight of the men in the elevator is more than 180 lbs, the elevator would be overloaded.

To determine the average weight beyond which the elevator would be considered overloaded, we can follow these steps:
Find the total weight limit for 33 occupants: 5940 lbs.

Divide the total weight limit by the number of occupants to find the average weight per person: 5940 / 33 = 180 lbs.
Now, we need to find the difference between the average weight per person (180 lbs) and the mean weight of the population (166 lbs): 180 - 166 = 14 lbs.
Since we have the difference and the standard deviation (26 lbs), we can now calculate the Z-score:

Z = (difference) / (standard deviation) = 14 / 26 ≈ 0.54.
The average weight beyond which the elevator would be considered overloaded is 180 lbs.

The corresponding Z-score for this weight is approximately 0.54.

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When a researcher uses the Pearson product-moment correlation, two highly correlated variables will appear on a scatter diagram as a tightly clustered group of points that form a linear pattern.

The scatter diagram is a visual representation of the correlation between two variables, where one variable is plotted on the x-axis, and the other variable is plotted on the y-axis. If the two variables have a high positive correlation, then the points on the scatter diagram will form a cluster that slopes upwards to the right.

On the other hand, if the two variables have a high negative correlation, then the points will form a cluster that slopes downwards to the right. The tighter the cluster of points, the higher the correlation between the variables.

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the total consumption of oil in the United States from January 1, 1980, to January 1, 1990 is approximately 596,533.7 billion barrels.

To find the total consumption of oil in the United States from January 1, 1980 to January 1, 1990, we need to integrate the given function[tex]C(t) = 27.08e^t[/tex]with respect to time t, over the interval [0, 10], where t is measured in years.

The integral of the function is:
∫[tex](27.08e^t) dt[/tex]

To solve this integral, we use the fact that the integral of[tex]e^t[/tex] is [tex]e^t[/tex]itself. Thus, we get:
27.08∫[tex]e^t dt = 27.08e^t[/tex]
Now, we need to evaluate the definite integral over the interval [0, 10]:
[tex]27.08e^t[/tex]| from 0 to 10 =[tex]27.08(e^{10} - e^0)[/tex]

As e^0 = 1, the expression simplifies to:
[tex]27.08(e^{10} - 1)[/tex]
Now, we can calculate the value of the expression:
[tex]27.08(e^{10} - 1) =27.08(22026.47 - 1) = 27.08 * 22025.47 =596533.7[/tex]
Thus, the total consumption of oil in the United States from January 1, 1980, to January 1, 1990 is approximately 596,533.7 billion barrels.

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Answer: 207.338[tex]cm^2[/tex] or 66[tex]\pi[/tex]

Step-by-step explanation:

Lateral Area of a cylinder : 2[tex]\pi[/tex](radius)(height)

Surface Area of a cylinder : Lateral Area + 2 (base area)

LA= 48[tex]\pi[/tex]

= 150.79

SA= 150.79 + (2([tex]\pi[/tex]([tex]3^2[/tex])

= 207.338 [tex]cm^2[/tex]

: )))

Answer: 9m

Step-by-step explanation:

there is one “m” in each and there is a 9 in both number. (9*2=18), (9*3=27)

The probability of selecting such a permutation is very low, only about 0.31%.

The probability of the event when we randomly select a permutation of the 26 lowercase letters of the English alphabet where 'm' immediately precedes 'n', which immediately precedes 'o' can be calculated as follows:

Firstly, we need to determine the total number of permutations of the 26 letters. Since there are 26 letters in the alphabet, there are 26! ways to arrange them.

Next, we need to determine the number of permutations where 'm' immediately precedes 'n', which immediately precedes 'o'. To do this, we can consider 'mno' as a single unit and then there are 24! ways to arrange the 24 units (23 individual letters and 1 unit of 'mno').

However, there are 3! ways to arrange 'mno' within the unit, so we need to multiply by 3!. Therefore, the total number of permutations where 'm' immediately precedes 'n', which immediately precedes 'o' is 24! x 3!.

Thus, the probability of randomly selecting a permutation where 'm' immediately precedes 'n', which immediately precedes 'o' is:

P = (24! x 3!) / 26!

P ≈ 0.0031 or 0.31%

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The chance that you will both win the raffle and win $500 is 0.0025, or 0.25%.

To find the chance of both winning the raffle and correctly guessing the organizer's favorite season, you need to multiply the probabilities of these two independent events.

Step 1: Determine the probability of winning the raffle.
The probability of winning the raffle is given as 0.01.

Step 2: Determine the probability of correctly guessing the favorite season.
The probability of correctly guessing the favorite season is given as 0.25.

Step 3: Multiply the probabilities of the two independent events.
To find the probability of both events happening, you multiply their probabilities: 0.01 (winning the raffle) * 0.25 (correctly guessing the favorite season).

0.01 * 0.25 = 0.0025

So, the chance that you will both win the raffle and win $500 is 0.0025, or 0.25%.

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The probability of both winning the raffle and correctly guessing the organizer's favorite season to win the $500 prize is 0.0025 or 0.25%.

To find the probability of both winning the raffle and guessing the organizer's favorite season correctly, you'll need to multiply the individual probabilities of each event.

Probability of winning the raffle: 0.01 (given in the question)
Probability of guessing the organizer's favorite season: 0.25 (given in the question)
Now, multiply these probabilities together:
0.01 * 0.25 = 0.0025.

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Since cosine is negative and a is in quadrant III, we know that sine is positive. We can use the Pythagorean identity to solve for sine:

sin^2(a) + cos^2(a) = 1

sin^2(a) + (-5/9)^2 = 1

sin^2(a) = 1 - (-5/9)^2

sin^2(a) = 1 - 25/81

sin^2(a) = 56/81

Taking the square root of both sides:

sin(a) = ±sqrt(56/81)

Since a is in quadrant III, sin(a) is positive. Therefore:

sin(a) = sqrt(56/81) = (2/3)sqrt(14)

Using the standard normal distribution, we find that approximately 73.8% of workers receive wages between $4.75 and $5.69 per hour. Using the inverse of the standard normal distribution, we find that the highest 5% of hourly wages are greater than approximately $6.09.

Using a standard normal distribution table or calculator with a mean of 5.25 and a standard deviation of 0.60, we can find that approximately 79.42% of workers receive wages between $4.75 and $5.69 per hour.

Using a standard normal distribution table or calculator, we can find the z-score corresponding to the highest 5% of wages, which is approximately 1.645.

Then, we can solve for x in the equation z = (x - μ) / σ, where z is the z-score, μ is the mean of 5.25, and σ is the standard deviation of 0.60. This gives us x = zσ + μ, which is approximately $6.09 per hour. Therefore, the highest 5% of hourly wages are greater than $6.09 per hour.

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Trigonometric functions are used to model relationships between angles and sides of a right triangle. They are particularly useful in solving problems that involve angles, distances, heights, and lengths that are difficult to measure directly.

For example, consider a problem that involves finding the height of a building. By measuring the length of the shadow cast by the building at a particular time of day, the angle of the sun's rays can be calculated using trigonometry. Once the angle is known, the height of the building can be determined using the tangent function.

In contrast, a non-trigonometric function may not be able to model the relationship between the given quantities in such problems, and may not provide an accurate solution. Therefore, when a problem involves angles or distances that are not directly measurable, trigonometric functions are typically the best tool for setting up and solving the problem.

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Answer:

Step-by-step explanation:

its a rectanlge

The length of arc PQ is 110 degrees

Knowing that the arc lengths are in degrees and the total for a circle is 360 degrees then we have the equation

8x - 10 + 6x + 10x + 10 = 360

To solve the equation 8x - 10 + 6x + 10x + 10 = 360 for x, we first need to simplify the left side of the equation by combining like terms:

8x + 6x + 10x - 10 + 10 = 24x

Now the equation becomes:

24x = 360

To solve for x, we need to isolate x on one side of the equation by dividing both sides by 24:

24x/24 = 360/24

x = 15

Therefore, the solution for x is 15.

Arc PQ = 8x - 10

= 8 * 15 - 10

= 110 degrees

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Answer:

19/14

Step-by-step explanation:

6/4+4/8

to solve that multiply the denominator on the left with 8 and the one on the right with 7 to make them equivalent and do the same for the numerator so now its 76/56 so now just simplified as much as possible will be 19/14

Answer:

Take your question 6/7 + 4/8, and find a common denominator. 7 and 8 both go in to 56, so set both denominators to 56.

The question now reads 6/56 + 4/56.

Multiply the numerator by the number of times the denominator was multiplied to get to the common denominator. 8 goes into 56 7 times, and 7 goes into 56 8 times.

Therefor:

7 x 8 = 56, and 8 x 7 = 56.

Now multiply  the numerator by the amount of times the denominator went into the common denominator.

8 x 6 = 48 and 7 x 4 = 28.

So 48/56 and 28/56.

We could then simplify, by dividing both sides by the same number.

12/14

7/14

Then add the numerators only.

That would give us 19/14

Hope that helps, :D.

Answer: To find the average rate of change of the function f(x) over the interval [9, 65], we can use the formula:

average rate of change = (f(b) - f(a)) / (b - a)

where a = 9, b = 65, f(a) = f(9) = 3√8 + 2, and f(b) = f(65) = 3√64 + 2.

Plugging in these values, we get:

average rate of change = (f(65) - f(9)) / (65 - 9)

average rate of change = (3√64 + 2 - 3√8 - 2) / 56

average rate of change = (3(4) + 2 - 3(2) - 2) / 56

average rate of change = (12 - 4) / 56

average rate of change = 8 / 56

average rate of change = 1 / 7

Therefore, the average rate of change of the function f(x) over the interval [9, 65] is 1/7.

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer: To find the average rate of change of the function f(x) over the interval [9, 65], we can use the formula:

average rate of change = (f(b) - f(a)) / (b - a)

where a = 9, b = 65, f(a) = f(9) = 3√8 + 2, and f(b) = f(65) = 3√64 + 2.

Plugging in these values, we get:

average rate of change = (f(65) - f(9)) / (65 - 9)

average rate of change = (3√64 + 2 - 3√8 - 2) / 56

average rate of change = (3(4) + 2 - 3(2) - 2) / 56

average rate of change = (12 - 4) / 56

average rate of change = 8 / 56

average rate of change = 1 / 7

Therefore, the average rate of change of the function f(x) over the interval [9, 65] is 1/7.

Below is the answer to the questions:

Q 12.

The prediction is that there were about 260 principals in attendance at the conference.

Q13.

The best prediction is that the college will have about 480 students who prefer cookies.

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Answer: 3 pieces

Step-by-step explanation:First, 6/8 can be converted into fourths by dividing the numerator and the denominator by 2 and we get 3/4. if we want 1/4 slices we divide 3/4 by 1/4 and get 3.

1) The common denominators of 2/3 and 7/9 are: 9, 18, 27, 36, 45, 54, 63, ...

2) The common denominators of 1/9 and 1/2 are: 18, 36, 54, 72, 90, 108, ...

To find the common denominators of 2/3 and 7/9, we need to find the least common multiple (LCM) of the denominators 3 and 9.

Prime factorization of 3: 3 = 3^1

Prime factorization of 9: 9 = 3^2

To find the LCM, we take the highest power of each prime factor that appears in either factorization. So, LCM(3, 9) = 3^2 = 9.

Therefore, the common denominators of 2/3 and 7/9 are all multiples of 9.

2) To find the common denominators of 1/9 and 1/2, we need to find the LCM of the denominators 9 and 2.

Prime factorization of 9: 9 = 3^2

Prime factorization of 2: 2 = 2^1

To find the LCM, we take the highest power of each prime factor that appears in either factorization. So, LCM(9, 2) = 2 x 3^2 = 18.

Therefore, the common denominators of 1/9 and 1/2 are all multiples of 18.

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The number of collections of 50 coins that can be chosen from this pile is: C(125, 25) = 177,100,565,136,000
This is a very large number, which shows that there are many possible collections of 50 coins that can be chosen from the pile.

If the pile contains only 25 quarters but at least 50 of each other kind of coin, then the total number of coins in the pile must be at least 50 + 50 + 50 = 150. Let's assume that there are 150 coins in the pile, including the 25 quarters.
To choose a collection of 50 coins from this pile, we need to exclude the 25 quarters and choose 25 coins from the remaining 125 coins. We can do this in C(125, 25) ways, which is the number of combinations of 25 items chosen from a set of 125 items.
Therefore, the number of collections of 50 coins that can be chosen from this pile is:
C(125, 25) = 177,100,565,136,000

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There are 351 possible collections of 50 coins that can be chosen, considering the given conditions.

To find the number of collections of 50 coins that can be chosen, we will consider the given conditions:
The pile contains only 25 quarters.

There are at least 50 of each other kind of coin (pennies, nickels, and dimes).
Now, let's break this down step by step:
Determine the minimum number of coins from each kind required to make a collection of 50 coins.
- 25 quarters (as it's the maximum available)
- The remaining 25 coins must be a combination of pennies, nickels, and dimes.
Find the different combinations of pennies, nickels, and dimes that can be chosen to make a collection of 50 coins.
- We need 25 more coins, so we can divide them into three groups:
 a) Pennies (P)
 b) Nickels (N)
 c) Dimes (D)
Calculate the combinations for the remaining 25 coins.
- Using the formula for combinations with repetitions: C(n+r-1, r) = C(n-1, r-1)
 Where n is the number of types of coins (3) and r is the number of remaining coins (25)
- C(3+25-1, 25) = C(27, 25) = 27! / (25! * 2!) = 351.

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The probability of at least 10 cities out 45 have a commute time greater than 30 minutes is 0.893 or 89.3%

Apply the complement rule.

Probability that at least one of the 10 cities you select will have a commute time greater than 30 minutes,

The complement of the event is 'none of the 10 cities have a commute time greater than 30 minutes'.

The probability of the complement event can be calculated by ,

Multiplying probabilities of selecting a city with a commute time less than or equal to 30 minutes for each of 10 selections.

P(none of 10 cities have a commute time > 30 minutes) = (35/45) x (35/45) x ... x (35/45) (10 times)

Because there are 35 cities out of the total 45 that have a commute time less than or equal to 30 minutes.

So the probability that at least one of the 10 cities has a commute time greater than 30 minutes is,

1 - P(none of the 10 cities have a commute time greater than 30 minutes)

= 1 - (35/45) x (35/45) x ... x (35/45) (10 times)

= 1 - 0.1073

= 0.8926

Therefore, the probability that at least one of the 10 cities you select will have a commute time greater than 30 minutes is 0.893 or 89.3%.

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The above question is incomplete, the complete question is:

If i randomly sample two cities from this group (consider these 45 the 'population' if you will) then what is the probability that at least one of the 10 cities i select will have a commute time greater than 30 minutes?

Answer:52.5

Step-by-step explanation:

Multiply

Answer:

The y-intercept is at (0, 8).

Answer: (0,8)

Step-by-step explanation: The line only touches the Y-axis Once and its on 8

The missing angle D is 55 degrees, and the lengths of DE and DF are approximately 8.40 units and 14.65 units, respectively. Therefore, the correct option is (B) m∠D = 55°, DE ≈ 8. 40 units, DF ≈ 14. 65 units

We can start by using the trigonometric ratios of the angles in a right triangle. In particular, we can use the tangent function to find the measure of angle D

tan(D) = DE / FE

tan(D) = DE / 12

We know that angle F is 35 degrees, so angle D must be

D = 90 - F

D = 90 - 35

D = 55 degrees

Now that we know the measure of angle D, we can use the sine and cosine functions to find the lengths of DE and DF, respectively. We know that

sin(F) = DE / DF

cos(F) = FE / DF

Substituting the given values

sin(35) = DE / DF

cos(35) = 12 / DF

Solving for DE and DF

DE = DF × sin(35)

DE = DF × 0.574

DE ≈ 0.574 × DF

DF = 12 / cos(35)

DF ≈ 14.65 units

DE ≈ 0.574 × 14.65

Multiply the numbers

DE ≈ 8.40 units

Therefore, the correct option is (B) m∠D = 55°, DE ≈ 8. 40 units, DF ≈ 14. 65 units

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The given question is incomplete, the complete question is:

Solve the given right triangle for its missing angle and side measures

A. m∠D = 55°, DE ≈ 4. 40 units, DF ≈ 13. 65 units

B. m∠D = 55°, DE ≈ 8. 40 units, DF ≈ 14. 65 units

C. m∠D = 35°, DE ≈ 8. 40 units, DF ≈ 13. 65 units

D. m∠D = 35°, DE ≈ 8. 40 units, DF ≈ 14. 65 units