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4 years ago

Can someone please help

Mathematics

2 answers:

AleksAgata [21]4 years ago

5 0

The answer is D. There is a one third change the book will be an adventure book because there are 10 adventure books and 30 books altogether. 10 / 30 = 1/3

defon4 years ago

4 0

B). One over thirty. That is the answer

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What is the magnification formula?

romanna [79]

Answer:

It is option A

Step-by-step explanation:

And the "suck it" guy is a loser :)

3 0

3 years ago

In 2002, there were 972 students enrolled at Oakview High School. Since then, the number of students has increased by 1.5% each

Katarina [22]

Answer:

The number of student increased each year by 986.58.

Step-by-step explanation:

Given : In 2002, there were 972 students enrolled at Oakview High School.

To find : The number of students has increased by 1.5% each year ?

Solution :

Using the exponential growth function,

$P(t)=P_o(1+r)^t$

Here, $P_o=972$ is the initial population

r=1.5%=0.015 is the rate

t=1 time

$P(t)=972(1+0.015)^1$

$P(t)=972(1.015)$

$P(t)=986.58$

i.e. the number of student increased each year by 986.58.

3 0

4 years ago

Which of the following is equivalent to 2^2/5?

pashok25 [27]

The answer is letter A since the mixed form of 7/5 is equal to 1 and 2/5.

3 0

3 years ago

Mary bowled six games this week. Her scores were 138,151,198,147,156, and 173. What was her average for those

erica [24]

The average of scores is 160.5

Step-by-step explanation:

Mean is the sum of all values divided by the number of values.

Given scores are:

138,151,198,147,156, and 173

The formula for average score is given by:

$Avg = \frac{Sum}{items}$

Putting values

$Avg = \frac{138+151+198+147+156+173}{6}\\Avg = \frac{963}{6}\\= 160.5$

The average of scores is 160.5

Keywords: Mean, Average

Learn more about mean at:

#LearnwithBrainly

4 0

4 years ago

Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviati

Anna71 [15]

Answer:

( 67.18, 68.82)

Step-by-step explanation:

Let $\mu$ be the true (population) mean of statistics exam scores. We have a large random sample of n = 36 scores with a sample mean (sample mean score) of $\bar{x} = 68$ . We know that the population standard deviation is $\sigma = 3$ . A pivotal quantity is $(\bar{x}-\mu)/(3/\sqrt{36}) = (\bar{x}-\mu)/(3/6) = (68-\mu)/(1/2)$ which is approximately normally distributed and $P(-z_{0.05}\leq(68-\mu)/(1/2)\leq z_{0.05}) = 0.90$ where $-z_{0.05} =-1.6449$ , and so, $P(-1.6449\leq(68-\mu)/(1/2)\leq 1.6449) = 0.90$ . Therefore the 90% confidence interval is (68-(1/2)(1.6449), 68+(1/2)(1.6449)), i.e., ( 67.18, 68.82)

8 0

3 years ago