F= 9/5C + 32, solve for F please.

To determine if a probability of contamination more than twice the mean of 14.4 is unusual, the standard deviation must be considered. In a normal distribution, occurrences more than two standard deviations from the mean are typically unusual.

Whether a probability of contamination more than twice the mean of 14.4 is unusual or not depends on the standard deviation of the data. In a typical normal distribution, values more than two standard deviations away from the mean could be considered unusual. Considering a mean of 14.4, if the standard deviation is small, then a value twice as large would likely be an outlier. However, if the standard deviation is large, such a value could fall within the realm of normal variation.

To ascertain if a value is unusual, one would compare it to the range defined by a certain number of standard deviations from the mean. A common rule of thumb is the empirical rule, which states that for a normally distributed dataset, roughly 68% of data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations.

In this context, if a probability of contamination is more than twice the mean (greater than 28.8), and assuming a normal distribution, it would be considered unusual if it falls outside of two standard deviations from the mean.

3 angles inside a triangle = 180 degrees

 since the 2 sides are equal

 angle B  and C are the same

180 - 50 = 130 degrees

130 / 2 =65

 Angle B is 65 degrees

1. The average rate of change is the slope of a line to find it divide the vertical change by the horizontal change

2. recursive is a formula to calculate the next term, depending on what the previous term was

 Explicit is a formula to calculate any term depending on the term number

Answer:

10 1/3 = 3 /31.      2 2/3 = 3/8.          5 1/3 = 3/16.

Step-by-step explanation:

i took the k12 test and got 100 percent hope this helps!!

Answer:

wow

Step-by-step explanation:

Answer:

Option B is right.

Step-by-step explanation:

Given that there are totally 9 books unread by Mrs. Reid.  OUt of these she wants to take only 3 books on the trip.

For books order does not count.

Hence combinations would be appropriate to calculate.

3 books out of 9 can be selected in

9C 3 ways

No of ways of selecting 3 books for the trip out of 9 = [tex]\frac{9(8)(7)}{1(2)(3)} =84[/tex]

There are just 84 ways of choosing 3 books to bring on the trip out of 9 unread books.

Option B is right.

The correct answer is 84.

To solve this problem, we can use the concept of combinations from combinatorics. A combination is a way of selecting items from a collection, such that the order of selection does not matter. In this case, Mrs. Reid wants to select 3 books out of 9 without regard to the order in which they are chosen.

The number of ways to choose 3 books from 9 is given by the combination formula:

[tex]\[ C(n, k) = \frac{n!}{k!(n-k)!} \][/tex]

where[tex]\( n \)[/tex] is the total number of items to choose from, [tex]\( k \)[/tex]is the number of items  to choose, and [tex]\( ! \)[/tex] denotes factorial, which is the product of all positive integers up to that number.

For Mrs. Reid's situation:

[tex]\[ n = 9 \] \[ k = 3 \][/tex]

So we plug these values into the formula:

[tex]\[ C(9, 3) = \frac{9!}{3!(9-3)!} \][/tex]

Calculating the factorials:

[tex]\[ 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \] \[ 3! = 3 \times 2 \times 1 \] \[ (9-3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]

Now we simplify the expression by canceling out the common terms in the numerator and the denominator:

[tex]\[ C(9, 3) = \frac{9 \times 8 \times 7 \times 6!}{3! \times 6!} \] \[ C(9, 3) = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \][/tex]

Divide out the common factors:

[tex]\[ C(9, 3) = \frac{9}{3} \times \frac{8}{2} \times \frac{7}{1} \] \[ C(9, 3) = 3 \times 4 \times 7 \][/tex]

Now we multiply the remaining numbers [tex]\[ C(9, 3) = 3 \times 4 \times 7 = 84 \][/tex]

Therefore, Mrs. Reid can choose 3 books to bring on the trip in 84 different ways.

We can use the points (-4, -1) and (0, -2) to solve.

Slope formula: y2-y1/x2-x1

= -2-(-1)/0-(-4)

= -1/4

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Answer:

B

Step-by-step explanation:

Answer:

[tex]2\sqrt{1+tan(x)} +C[/tex]

Step-by-step explanation:

To start solving this you need to use substitution. I let u = 1+tan(x). Next you need to find du/dx, which is sec^2(x) using trigonometric properties. Solve for dx and get dx = du / sec^2(x). Next put the new dx back in. This gives you integral sec^2(x) / sqrt u * du / sec^2(x). The sec^2(x) cancels and the new expression is integral 1/sqrt u * du, which can be simplified to integral u^-1/2 * du. You then take the integral and get 2u^1/2. Lastly, substitute the original u back in and get 2 sqrt 1+tan(x) + C.

Given that the certain type of spore colony having 50 spores and a rate of growth of 3 times in 24 hour or one day period, it would grow to (3 times 50) 150 spores at the end of the first day, (3 times 150) 450 in the 2nd day, (3 times 450) 1350 in the third day, and (3 times 1350) 4050 spores at the end of the fourth day.

The spore colony, which triples in size each day, will contain 4050 spores at the end of the 4th day starting from an initial count of 50 spores.

This question involves exponential growth, where the size of the spore colony triples every day.

To solve this, we will calculate the number of spores at the end of each day:

Initial spores= 50

End of Day 1,
50 * 3 = 150

End of Day 2,
150 * 3 = 450

End of Day 3,
450 * 3 = 1350

End of Day 4,
1350 * 3 = 4050

Therefore, at the end of the 4th day, the spore colony would contain 4050 spores.

Our three consecutive integers are 37, 38, and 39.

Work is provided in the image attached.

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