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3 years ago

Brandon has 32 stamps. he wants to display the stamps in rows,with the same n

1 answer:

3 0

Okay, let's work this out...
What we know:
-32 stamps in all
- rows (horizontal)
- same # in each
What we "want to know" :
- # of combinations (different)
Problem Solving :
This is actually very easy its just the words than get ya!
1st : we need to figure out the factors of 32...
In other words, we need to figure out _x_=32 and how many different combinations and ways there are!
Note:(* means multiplication)
#1: What are the factors of 32?
32: 1*32 , 2*16 , 4*8
32: 32*1 , 16*2 , 8*4
The factors (not including 1 are 2,4,8,16)
Now, as you can see, there are 4 ways to get 32 as shown first.
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That ¦ is 1 way with 16 in 2 rows. Basic multiplication, 16*2=32 or 16+16=32.
But, this is also a way,
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Now there should be 2 in each row and 16 rows. Again 2*16=32 or 2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2=32
That's two ways so far.
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Another way which is 4 rows with 8 in each.
4*8=32 or 8+8+8+8=32
But, this is also a way,
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Now that is 8 rows with 4 in each. 8*4=32 or 4+4+4+4+4+4+4+4=32
That was our fourth way.
Again NOT including 1. If you include 1 then there will be 6 ways but aside from that there are 4 ways.
I hope that helped I worked hard typing this all for you. Any questions just ask!

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3 years ago

A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

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3 years ago

What is 47 minutes before 6:57

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2 years ago

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chubhunter [2.5K]

Answer:

After 2009 the quality of the forests in this country improving.

Step-by-step explanation:

Consider the provided function.

$f(t) = \frac{1}{3}t^3-\frac{5}{2}t^2 + 80$

$f'(t) = t^2-5t$

Substitute f'(t)=0.

$t^2-5t=0$

$t(t-5)=0$

$t=0\ or\ t=5$

It is given that (0 ≤ t ≤ 10)

The test interval are (0,5) and (5,10)

For (0 < t < 5) the value of $f'(t)$

For (5 < t < 10) the value of $f'(t) >0$

Hence, the function is increasing from (5,10).

t = 0 corresponds to 2004.

Therefore, after 2009 the quality of the forests in this country improving.

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