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Olegator [25]
3 years ago
9

A circular, rotating, serving tray has 7 different desserts around its circumference. In how many different ways can all the des

serts be arranged on the circular tray?
Mathematics
1 answer:
Romashka [77]3 years ago
4 0
If there are n different things in a circular arrangement, the number of ways of arranging them is (n - 1)!
The number of ways the desserts can be arranged on the circular tray is:
(7 - 1)! = 6 * 5 * 4 * 3 * 2 * 1 = 720 ways
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The two way frequency table shows what a group of students did during the summer. Which statement is true related to the table?
ratelena [41]

Answer:

That statement is true which relates to what activity was taken by most of the students of different categories in that particular group during the summers.

Step-by-step explanation:

A two-way frequency table (likewise called a possibility table) is a valuable instrument for inspecting connections between downright factors. The passages in the cells of a two-way table can be frequency checks or relative frequencies (simply like a single direction table). Two-way frequency tables show what number of information focuses fit in every classification.  

Assume you direct a review where you pose every individual two inquiries. When you have wrapped up the overview, you will have two bits of information from every individual. At whatever point you have two bits of information from every individual, you can arrange the information into a two-way frequency table.

8 0
2 years ago
4(b-6)+19 simplified
Firlakuza [10]

Answer:

4(b-6)+19

4(b) +4(-6) +19 distributive property

4b-24+19 combine like terms

4b-5

4 0
2 years ago
A scientist is studying the growth of a particular species of plant. He writes the following equation to show the height of the
gtnhenbr [62]

At day 0 the height of the plant = 12 cm

Calculate how many days  to get to 16.13 cm:-
16.13  = 12(1.03)^n

1.03^n = 16.13 / 12 = 1.34417
n = ln 1.34417 / ln 1.03  = 10  days

I would think that a reasnable domain would be   0 =< n <= 100
5 0
2 years ago
PLEASE HELP!
Aleks04 [339]

Answer:

Option c:

0.25 \pm 1.645\sqrt{\frac{0.25*(1-0.25)}{90}}

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

For this problem, we have that:

You have access to first year enrolment records and you decide to randomly sample 119 of those records. You find that 89 of those sampled went on to complete their degree. This means that n = 119, \pi = \frac{89}{119} = 0.7478.

90% confidence level

So \alpha = 0.1, z is the value of Z that has a p-value of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

Sample of 90 people with a sample proportion of 0.25

This means that n = 90, p = 0.25.

Confidence interval:

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

0.25 \pm 1.645\sqrt{\frac{0.25*(1-0.25)}{90}}

Which is option c.

6 0
2 years ago
Like I don't get I can someone help me please
deff fn [24]

Answer:

The answer is 9

Step-by-step explanation:

It is 9 because you have to simplify 27 and 18 so for 18 its 6 and for 27 it's 9

3 0
2 years ago
Read 2 more answers
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