There are 365,493 blue pens and 549,384 black pens how many more black pens are there then blue pens

The expression simplifies using multiple properties of exponents, including the power of a product, properties of exponents for multiplication and division, the zero exponent property, and squaring.

The expression you're asking about applies multiple properties of exponents to simplify the given expression. To start, applying the power of a product property, it simplifies [tex](4*9)^5[/tex] to [tex]4^5*9^5[/tex]. Following that, the property of exponents for division is used to combine [tex]4^{-5[/tex] and [tex]4^5[/tex], leading to [tex]4^0[/tex], as well as [tex]9^5[/tex] and [tex]9^{-3[/tex], resulting in [tex]9^2[/tex]. The zero exponent property translates [tex]4^0[/tex]into 1. Lastly, the square of 9 gives the end result of 81.

In summary, these are the properties applied: power of a product, properties of exponents for multiplication and division, zero exponent property, and squaring.

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In a 4-year period, expect around 2 loads (Poisson model). Over 11 loads is likely (80% chance). No loads for 5.8 years or more guarantees under 30% chance.

In a Poisson process with a mean time between occurrences of λ = 0.4 years, we can answer your questions as follows:

**(a) Expected number of loads in 4 years:**

The expected number of events in a Poisson process is the product of the mean occurrence rate (λ) and the time interval (t). Therefore, in a 4-year period, you can expect:

Expected loads = λ * t = 0.4 years/load * 4 years ≈ 1.6 loads

However, since the Poisson process deals with discrete events, the expected value doesn't necessarily translate to an exact number of occurrences. It represents the average number of loads you would expect to see over many repetitions of the same 4-year period.

**(b) Probability of more than 11 loads in 4 years:**

To calculate the probability of more than 11 loads, we need to find the probability of 12 or more loads. In a Poisson process, the probability of k events occurring in a time interval t is given by:

P(k events) = (e^(-λt) * (λt)^k) / k!

For k = 12 and λt = 4 * 0.4 = 1.6:

P(more than 11 loads) = 1 - P(0 loads) - P(1 load) - ... - P(11 loads)

≈ 1 - (e^(-1.6) * 1.6^0) / 0! - (e^(-1.6) * 1.6^1) / 1! - ... - (e^(-1.6) * 1.6^11) / 11!

≈ 1 - 0.1995 ≈ 0.8005

Therefore, the probability of having more than 11 loads in a 4-year period is approximately 0.801 (rounded to three decimal places).

**(c) Time period for at most 0.3 probability of no loads:**

We want to find the time period (t) where the probability of no loads (P(0 loads)) is at most 0.3. In other words, we need to solve:

P(0 loads) = e^(-λt) ≤ 0.3

Taking the natural logarithm of both sides and rearranging:

t ≥ -ln(0.3) / λ ≈ 2.303 / 0.4 ≈ 5.76 years

Therefore, the time period must be at least 5.76 years (rounded to four decimal places) for the probability of no loads to be at most 0.3.

The probable question may be:

An article suggests that a Poisson process can be used to represent the occurrence of structural loads over time. Suppose the mean time between occurrences of loads is 0.4 year.

(a) How many loads can be expected to occur during a 4-year period?

(b) What is the probability that more than eleven loads occur during a 4-year period? (Round your answer to three decimal places.)

(c) How long must a time period be so that the probability of no loads occurring during that period is at most 0.3? (Round your answer to four decimal places.)

To answer problems like this you have to use binomial:

P (x > 1) = 1 – p (0 < x < 1) > .7 

So:

1 – p (0) – p (1) > .7 

1 – (3/ 4) ^n – (3/ 4) ^n (n – 1 ) (1/ 4) > .7 

Therefore n > 5.185, and the smallest value of n so that we can satisfy the given condition is 6 (rounded up)

Final answer:

Yes, 0.91919191 can be written as a fraction. The fraction form is 91919191/9999999.

Explanation:

Yes, 0.91919191 can be written as a fraction. To convert a repeating decimal to a fraction, we can use the following steps:

Therefore, 0.91919191 can be written as the fraction 91919191/9999999.

The gas mileage of a car that drives 283 miles on 12.3 gallons of gas is approximately 23.0 miles per gallon, when rounded to the nearest tenth.

Gas mileage is determined by dividing the total miles driven by the gallons of gas consumed. In this scenario, 283 miles were covered using 12.3 gallons of gas. The calculation yields approximately 22.97 miles per gallon (mpg), but rounding to the nearest tenth results in 23.0 mpg. This figure represents the efficiency of the vehicle in terms of fuel consumption, offering a quick and easy measure for assessing its economic and environmental impact. Monitoring gas mileage is crucial for budgeting and understanding the ecological footprint of transportation choices, promoting conscientious and efficient use of fuel resources.

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Answer:  SST for the data set is 32.

Step-by-step explanation:

Since we have given that

Coefficient of determination = R² = 0.25

Value of SSE = 24

We need to find SST.

As we know the formula for "Coefficient of determination":

[tex]R^2=1-\dfrac{SSE}{SST}[/tex]

We will put the above mentioned value:

[tex]0.25=1-\dfrac{24}{SST}\\\\0.25-1=-\dfrac{24}{SST}\\\\-0.75=-\dfrac{24}{SST}\\\\SST=\dfrac{24}{0.75}\\\\SST=32[/tex]

Hence, SST for the data set is 32.

To find the distance, multiply the speed by the time for each leg of the trip and add them together to get the total distance of 735 miles.

To find the distance to the parcel of land, we can use the formula Distance = Speed x Time. Since the pilot flew out and back, we can calculate the distance for each leg of the trip and then add them together.

Distance going out = Speed x Time = 120 mph x (7/2) hours = 420 miles

Distance coming back = Speed x Time = 90 mph x (7/2) hours = 315 miles

Total distance = Distance going out + Distance coming back = 420 miles + 315 miles = 735 miles

The probability of rolling a 5 on the green die and a 3 on the red die, or a 3 on the green die and a 5 on the red die is calculated by summing their individual probabilities, leading to a final probability of 1/18.

To solve this probability question, we will use the fundamental counting principle and basic probability rules. We have two six-sided dice, one green and one red. The question specifies two separate events:

The green die shows a 5 and the red die shows a 3.

The green die shows a 3 and the red die shows a 5.

Since each face on a fair six-sided die has an equal chance of appearing, the probability of getting any specific number is 1/6. When we roll two dice, the probability of any specific outcome is (1/6) x (1/6), which simplifies to 1/36.

Since there are two independent events that satisfy the question, we can add the probabilities of each event occurring:

Probability of a 5 on green die and 3 on red die = 1/36

Probability of a 3 on green die and 5 on red die = 1/36

Adding them up, we get 1/36 + 1/36 = 2/36, which simplifies to 1/18. Therefore, the probability of getting either a (5 on the green die and 3 on the red die) or (3 on the green die and 5 on the red die) is 1/18.

Answer:

Option C is correct

(2, 1) ordered pair comes from the table

Step-by-step explanation:

Ordered pair  states that an ordered pair contains the coordinate of points in the coordinate plane in the form of (x, y)

Given the table:

X Y  

2 1  

3 4

4 4

5 2

Ordered pairs for this table are:

From the definition, we have;

(2, 1), (3, 4), (4, 4) and (5, 2)

Therefore, the only ordered pairs comes from the table is, (2, 1)

Answer: Number of blueberries picked by Nico =78

Number of blueberries picked by Kai =858

Step-by-step explanation:

Given : Kai picked 11 times as many blueberries as Nico.  

Let , Number of blueberries picked by Nico =x

Then, Number of blueberries picked by Kai = 11x

Together, they picked 936 blueberries.

[tex]\Rightarrow\ x+11x=936\\\\\Rightarrow\ 12x=936\\\\\Rightarrow\ x=\dfrac{936}{12}=78[/tex]

Now, Number of blueberries picked by Nico =78

Then, Number of blueberries picked by Kai = 11(78) =858