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2 years ago

What is the sum of StartRoot negative 2 EndRoot and StartRoot negative 18 EndRoot?

Mathematics

1 answer:

Darya [45]2 years ago

3 0

Answer:

$4 \sqrt{2} i$

Step-by-step explanation:

We want to find the sum of

$\sqrt{ - 2} + \sqrt{ - 18}$

We can rewrite this as

$\sqrt{ 2 \times - 1} + \sqrt{ 18 \times - 1}$

This becomes;

$\sqrt{2} \times \sqrt{ - 1} + \sqrt{18} \times \sqrt{ - 1}$

Recall that;

$i = \sqrt{ - 1}$

This implies that;

$\sqrt{2} i + 3 \sqrt{2} i$

Combine like terms:

$4 \sqrt{2} i$

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Which explains how to find the quotient of the division below? - 3 1/3 divided by 4/9 Write Negative 3 and one-third as Negative

makvit [3.9K]

Answer:

its B

Step-by-step explanation:

cuz -3 1/3 would be turned into -10/3 by multiplying 3*3 which will give u 9 and adding 1 which is 10 then u put the 10 on top and keep the 3 as bottom.

and then flip the 4/9 which will turn into 9/4 so now u can multiple intead of divide

-10/3 * 9/4 and then youll get ur answer.

4 0

2 years ago

Read 2 more answers

Suppose 1000 people enter a chess tournament. Use a rooted tree model of the tournament to determine how many games must be play

Vlada [557]

Answer: 999 games

Step-by-step explanation:

There are many ways to illustrate the rooted tree model to calculate the number of games that must be played until only one player is left who has not lost.

We could go about this manually. Though this would be somewhat tedious, I have done it and attached it to this answer. Note that when the number of players is odd, an extra game has to be played to ensure that all entrants at that round of the tournament have played at least one game at that round. Note that there is no limit on the number of games a player can play; the only condition is that a player is eliminated once the player loses.

The sum of the figures in the third column is 999.

We could also use the formula for rooted trees to calculate the number of games that would be played.

$i=\frac{l - 1}{m - 1}$

where i is the number of "internal nodes," which represents the number of games played for an "<em>m</em>-ary" tree, which is the number of players involved in each game and l is known as "the number of leaves," in this case, the number of players.

The number of players is 1000 and each game involves 2 players. Therefore, the number of games played, i, is given by

$i=\frac{l - 1}{m - 1} = i=\frac{1000 - 1}{2 - 1} = \frac{999}{1} =999$

7 0

2 years ago

Crossett Trucking Company claims that the mean weight of its delivery trucks when they are fully loaded is 6,000 pounds and the

solniwko [45]

Answer:

(5953.52,6046.49)

Step-by-step explanation:

We are given the following in the question:

Mean, $\mu$ = 6,000 pounds

Sample size, n = 40

Alpha, α = 0.05

Standard deviation, σ = 150 pounds

95% Confidence interval:

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

$6000 \pm 1.96(\dfrac{150}{\sqrt{40}} )\\\\ = 6000 \pm 46.4854=\\(5953.5146,6046.4854)\approx (5953.52,6046.49)$

are the limits within which 95% of the sample means occur.

6 0

2 years ago

The product of 10 and the difference of a number x and 14 is 90

Ostrovityanka [42]

Write in numbers not words

3 0

2 years ago

How do you do box and whisker plots?

Juliette [100K]

Answer:

Find the five number summary of the data:

Minimum = x

Quartile 1 = x

Quartile 2 (median) = x

Quartile 3 = x

Maximum = x

Then, make the plot based on that information. If you are provided with a list of data, you should be able to use a calculator to find out that information.

5 0

2 years ago