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3 years ago

How many modes does the following data set have?

Mathematics

2 answers:

lesya [120]3 years ago

4 0

Answer:2

Step-by-step explanation:

25 and 4 both occur 4 times

Ede4ka [16]3 years ago

4 0

Answer: 2

Step-by-step explanation:

4 and 25 bcoz they appear most of the time.

You might be interested in

If four times a certain number, increased by 6 is equal to 94, what is the number?

aniked [119]

"4x + 6 = 94" is the equation among the following choices given in the question that could be used to solve the problem. The correct option among all the options that are given in the question is the first option. I hope that this is the answer that you were looking for and it has actually come to your help.

7 0

4 years ago

To celebrate the first day of a leap year, I taught my dog to jump through a hoop. It was a Sunday. When he taught me the same t

juin [17]

Answer:

Tuesday

Step-by-step explanation:

364/7 = 52, meaning that 364 days later will also be a Sunday. 2 days after that (366 as it is a leap year) would be a Tuesday.

8 0

4 years ago

What is the degree of each polynomial?

GREYUIT [131]

Answer:

A- 4

B- 5

C- 8

D- 4

E- 2

F- 7

I'm 99% sure these are right!! I hope this helps :(

3 0

3 years ago

Suppose f(x) = x^2 and g(x) = (1/3x)^2. Which statement best compares the graph of g(x) with the graph of f(x)?

miv72 [106K]

Since it would be (1/3) squared, it would be a vertical compression of 9. since the x is not the output, it would make a horizontal stretch factor of 3.

5 0

3 years ago

A certain class has 20 students, and meets on Mondays and Wednesdays in a classroom with exactly 20 seats. In a certain week, ev

Olin [163]

Answer:

The probability that no one sits in the same seat on both days of that week is given by, $P(\cap^{20}_{i=1}A_i^c)=\frac{1}{e}$

Step-by-step explanation:

Given : A certain class has 20 students, and meets on Mondays and Wednesdays in a classroom with exactly 20 seats. In a certain week, everyone in the class attend both days. On both days, the students choose their seats completely randomly (with one student per seat).

To find : The probability that no one sits in the same seat on both days of that week ?

Solution :

Let $A_i$ be the i-th student sits on seat which he has been sitting on Monday.

According to question,

We have to calculate $P(\cap^{20}_{i=1}A_i^c)$

Applying inclusion exclusion formula,

$P(\cap^{20}_{i=1}A_i^c)=1-P(\cap^{4}_{i=1}A_i)$

$P(\cap^{20}_{i=1}A_i^c)=1-P(A_1)+...+P(A_{20})-P(A_1\cap A_2)+...+P(A_{19}\cap A_{20})+P(A_1\cap A_2\cap A_3)+...+P(A_{18}\cap A_{19}\cap A_{20})....-P(A_1\cap A_2...\cap A_{20})$

Using symmetry,

$P(\cap^{20}_{i=1}A_i^c)=1-\sum^{20}_{k=1}(-1)^{k+1}\binom{20}{k}P(A_1\cap ...\cap A_k)$

$P(\cap^{20}_{i=1}A_i^c)=1-\sum^{20}_{k=1}(-1)^{k+1}\binom{20}{k}\frac{(20-k)!}{20!}$

$P(\cap^{20}_{i=1}A_i^c)=1+\sum^{20}_{k=1}(-1)^{k}\frac{1}{k!}$

$P(\cap^{20}_{i=1}A_i^c)=\sum^{20}_{k=0}(-1)^{k}\frac{1}{k!}$

$P(\cap^{20}_{i=1}A_i^c)=\frac{1}{e}$

Therefore, The probability that no one sits in the same seat on both days of that week is given by, $P(\cap^{20}_{i=1}A_i^c)=\frac{1}{e}$

8 0

3 years ago