Write 760% as a decimal and as a simplified fraction

The outliers are 0.8, 1.1, 10.2, and 10.9. The rest are not outliers.

A value is typically considered an outlier if it is less than Q1 - 1.5 [tex]\times[/tex] IQR or greater than Q3 + 1.5 [tex]\times[/tex] IQR.

First, we need to arrange the data in ascending order:

0.8, 1.1, 4.9, 5.2, 5.8, 5.9, 6.1, 6.1, 7.4, 10.2, 10.9

Next, we find the first quartile (Q1), which is the median of the first half of the data:

(4.9 + 5.2) / 2 = 5.05

We find the third quartile (Q3), which is the median of the second half of the data:

(6.1 + 7.4) / 2 = 6.75

Now, we calculate the interquartile range (IQR):

IQR = Q3 - Q1 = 6.75 - 5.05 = 1.7

The lower bound for outliers is:

Q1 - 1.5 [tex]\times[/tex] IQR = 5.05 - 1.5 [tex]\times[/tex] 1.7 = 5.05 - 2.55 = 2.5

The upper bound for outliers is:

Q3 + 1.5 [tex]\times[/tex] IQR = 6.75 + 1.5 [tex]\times[/tex] 1.7 = 6.75 + 2.55 = 9.3

Now we can determine which values are outliers:

0.8 is less than the lower bound (2.5), so it is an outlier.

1.1 is less than the lower bound (2.5), so it is an outlier.

10.2 is greater than the upper bound (9.3), so it is an outlier.

10.9 is greater than the upper bound (9.3), so it is an outlier.

The remaining values (4.9, 5.2, 5.8, 5.9, 6.1, 6.1, 7.4) are not outliers as they fall within the range of Q1 - 1.5 [tex]\times[/tex] IQR and Q3 + 1.5 [tex]\times[/tex] IQR.

Answer:  0.5

Step-by-step explanation:

Given :  The number of watches produced every hour is normally distributed with a mean of 500 and a standard deviation of 100.

i.e. [tex]\mu = 500\text{ and } \sigma= 100[/tex]

Let x be the number of watches produced every hour.

Then, the probability that in a randomly selected hour the number of watches produced is greater than 500 will be :

[tex]P(x>500)=1-P(x\leq500)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{500-500}{100})\\\\=1-P(z\leq0)\ \ [\because\ z=\dfrac{x-\mu}{\sigma}]\\\\=1-0.5\ \ [\text{ By z-table}]\\\\=1-0.5=0.5[/tex]

Hence, the probability that in a randomly selected hour the number of watches produced is greater than 500 =0.5.

Answer:D.  y<5x+30/x

y>6x+25/x

Step-by-step explanation:

got right on edmentum

The probability that the number of likely voters who think that it is too easy to vote is​:

a) exactly​ three = 0.17422

(b) at least 4 = 0.70533

(c) less than eight = 0.96406

How to use binomial probability theorem?

The general formula for binomial probability theorem is:

P(X = x) = ⁿCₓ * pˣ * (1 - p)⁽ⁿ ⁻ ˣ⁾

Where:

n = 12

p = 0.37

(a) exactly three,

P(x = 3) = ¹²C₃ * 0.37³ * 0.93⁹

= 0.17422

(b) at least 4,

P(4 ≤ x ≤ 12) = P(4) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10) + P(11) + P(12)

= 0.70533

(c) less than eight.

P(x < 8) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7)

= 0.96406

The probability that all six values on the first die and all six values on the second die occur once in the six rolls of the two dice, given the constraint that none of the forbidden pairs occur, is approximately 0.054%.

Here's the breakdown of the calculation:

Total possible outcomes:

Each die has 6 possible outcomes, so for 6 rolls, there are 6^6 = 46,656 possible combinations of rolls.

Outcomes with forbidden pairs:

We need to subtract the outcomes that contain any of the forbidden pairs.

There are 7 forbidden pairs, and each pair can occur in 6 different roll positions (e.g., (1, 5) could occur in the first roll, second roll, etc.).

However, we need to account for duplicates, as some of the forbidden pairs overlap in terms of the numbers involved (e.g., (5, 3) and (5, 5) both involve a 5 on the first die).

After careful calculation, considering the overlaps, there are 54 unique combinations with forbidden pairs.

Favorable outcomes:

We want all 6 values on each die to occur once.

There are 6! (6 factorial) = 720 ways to arrange the 6 values on the first die, and 720 ways to arrange the 6 values on the second die.

However, we don't care about the order within each die, so we divide by 6! twice to account for overcounting.

This leaves us with 720^2 / (6!)^2 = 1 favorable outcome.

Probability:

Probability = Favorable outcomes / Total possible outcomes

Probability = 1 / (46,656 - 54) ≈ 0.0005401235

Therefore, the probability of this specific event occurring is approximately 0.0005401235, or about 0.054%.

If x < 18 then you can subtract both sides by 18 and you would get x - 18 < 18 - 18 = 0

So that would be x - 18 < 0.

Hope this helps.

Final answer:

For the given data, the median is 85, the first quartile is 37, the third quartile is 90.5, the interquartile range (IQR) is 53.5, and the range is 72.

Explanation:

The median in a dataset is the middle value when the values are arranged in numerical order. For the given data 27, 33, 34, 35, 37, 39, 40, 41, 54, 75, 84, 75, 90, 87, 99, 91, 85, 88, 76, 92, 94, after arranging them in ascending order, the median (Q2) is the value positioned at the center of the dataset. Since there are 21 values, the median is the 11th value, which is 85. The first quartile (Q1) is the median of the first half (lower half) of the dataset, which is the value in the 5.5th position, so we average the 5th and 6th values to get 37. Similarly, the third quartile (Q3) is the median of the second half (upper half) of the dataset, which would be the average of the 16th and 17th values, yielding 90.5. The interquartile range (IQR) is the difference between the third quartile and the first quartile, so IQR = Q3 - Q1 = 90.5 - 37 = 53.5. The range of the dataset is the difference between the maximum and minimum value which is 99 - 27 = 72.

Answer:

actually its b and d

Step-by-step explanation:

according to khan academy

To round 1,563,385 to the nearest million, we look at the hundred-thousands digit (5), and since it's 5 or greater, we round up to 2,000,000.

To round off 1,563,385 to the nearest million, look at the hundred-thousands digit, which is 5 in this case.

If it is 5 or greater, we round up; if it is less than 5, we round down.

Since our hundred-thousands digit is 5, we round up, so 1,563,385 rounded to the nearest million is 2,000,000.