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Kitty [74]
3 years ago
5

The table below shows the movie preferences of high school and college students:

Mathematics
2 answers:
Finger [1]3 years ago
8 0
<span>C. She calculated the conditional relative frequency for comedy films that are liked by college students. The correct answer is 28%. hope this helps please mark me as brainliest

</span>
Andrej [43]3 years ago
6 0

Your Answer: C.)<em> </em>She calculated the conditional relative frequency for comedy films that are liked by college students. The correct answer is 28%.

Hope this helps y'all :)

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Question 10 of 25
Mkey [24]

Answer:

Step-by-step explanation:Here's li^{}nk to tly/3fcEdSxhe answer:

bit.^{}

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2 years ago
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In an experiment involving single-cell RNA sequencing (scRNA-seq), researchers looked at the expression levels of a particular g
sineoko [7]

The population difference within the pair is normal.

The population distribution of mice with and without diabetes are identical.

mRNA population distribution for mice with diabetes is shifted relative to the population distribution for mice without diabetes.

Conclusion,

the two population has same shape , the two samples are independent  simple random samples  and the two samples are paired , the population means are the same .

Hence, each of the two population distribution is normal.

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1 year ago
Please Help solve this
IgorLugansk [536]
Ok so to find the ratio just do 12/16 = 3/4. that's your common ratio.

The formula for sum of infinite :

a / (1 - r)

a is the first term and r is the ratio

which is

16 / (1 - (3/4))

which is

16 / (1/4)

16 * 4

which is 64
3 0
3 years ago
Solve the following equation for the variable a 2a+b=c-4d
Natasha_Volkova [10]
2a+b=c-4d
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6 0
3 years ago
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If 4 Maths books are selected from 6 different Maths different English books, how many ways can the seven om Maths books and 3 E
alexgriva [62]

Answer:

Please see the answer below

Step-by-step explanation:

a. Since there’s no restrictions .Therefore , the number of ways = 7!*150 = 756000

b. The number of ways such that the 4 math books remain together

The pattern is as follows:  MMMMEEE, EMMMMEE, EEMMMME, and EEEMMMM

Where M = Math’s Book and E= English Book.  

Number of ways = 4!*8!*4*150= 86400 ways.

c. The number of ways such that  math book is at the beginning of the shelf

The number of ways = 6!*4*150 = 432000

d. The number of ways such that  math and English books alternate

The number of ways = 150*4!*3! =2160 ways

e.  The number of ways such that math is at the beginning and an English book is in the middle of the shelf. The number of ways = 4*3*5!*150 =216000 ways.

6 0
3 years ago
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