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torisob [31]
3 years ago
13

What is the mean of the values in the dot plot? Enter your answer in the box. A number line ranging from 15 to 25 with one dot o

ver 16, four dots over 17, four dots over 19, one dot over 21, two dots over 22, one dot over 23, and one dot over 25.
Mathematics
2 answers:
solmaris [256]3 years ago
4 0
The mean of a dataset is the all members of the dataset added together and then divided by the number of members in the set. From your description of the number line, I've created a list of members.

16, 17, 17, 17, 17, 19, 19, 19, 19, 21, 22, 22, 23, 25

Adding all members of this set up produces the value 273. There are 14 members in the set, so 273 / 14 = 19.5. Thus, the mean of the set is 19.5

igomit [66]3 years ago
4 0

Answer:

19.5

Step-by-step explanation: I took the test.

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Four friends were born in consecutive years. The sum of their ages is 62. What is the age of the oldest friend?
marin [14]
The ages range from 62, 63, 64, 65 oldest friend age is 65 
7 0
4 years ago
Find the value of 2x-yi fx+y=8and4x-y=22​
Alexeev081 [22]

Answer:

10

Step-by-step explanation:

Given

x+y=8\\ \\4x-y=22

Add these two equations:

x+y+4x-y=8+22\\ \\5x=30\\ \\x=6

Substitute it into the first equation:

6+y=8\\ \\y=8-6\\ \\y=2

Then

2x-y=2\cdot 6-2=12-2=10

5 0
3 years ago
A.Find a formula for
snow_lady [41]

Answer:

a) \frac{n}{n+1}

b) Proof in explanation.

Step-by-step explanation:

a)

\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{n(n+1)}.

So let's look at the last term for a minute:

\frac{1}{n(n+1)}

Let's use partial fractions to see if we can find a way to write this so it is more useful to us.

\frac{1}{n(n+1)}=\frac{A}{n}+\frac{B}{n+1}

Multiply both sides by n(n+1):

1=A(n+1)+Bn

Distribute:

1=An+A+Bn

Reorder:

1=An+Bn+A

Factor:

1=n(A+B)+A

This implies A=1 and A+B=0 which further implies that B=-1.

This means we are saying that:

\frac{1}{n(n+1)} can be written as \frac{1}{n}+\frac{-1}{n+1}

We can check by combing the fractions:

\frac{n+1}{n(n+1)}+\frac{-n}{n(n+1)}

\frac{n+1-n}{n(n+1)}

\frac{1}{n(n+1)}

So it does check out.

So let's rewrite our whole expression given to us using this:

(\frac{1}{1}+\frac{-1}{2})+(\frac{1}{2}+\frac{-1}{3})+(\frac{1}{3}+\frac{-1}{4})+\cdots +(\frac{1}{n}+\frac{-1}{n+1})

We should see that all the terms in between the first and last are being zeroed out.

That is, this sum is equal to:

\frac{1}{1}+\frac{-1}{n+1}

Multiply first fraction by (n+1)/(n+1) so we can combine the fractions:

\frac{n+1}{n+1}+\frac{-1}{n+1}

Combine fractions:

\frac{n}{n+1}

b)

Proof:

Let's see what happens when n=1.

Original expression gives us \frac{1}{1 \cdot 2}=\frac{1}{2}.

The expression we came up with gives us \frac{1}{1+1}=\frac{1}{2}.

So it is true for the base case.

Let's assume our expression and the expression given is true for some integer k greater than 1.

We want to now show it is true for integer k+1.

So under our assumption we have:

\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots \frac{1}{k(k+1)}=\frac{k}{k+1}

So let's add the (k+1)th term of the given series on both sides:

\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots \frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}

(Now we are just playing with right hand side to see if we can put it in the form our solution which be if we can \frac{k+1}{k+2}.)

I'm going to find a common denominator which will be (k+1)(k+2):

\frac{k}{k+1} \cdot \frac{k+2}{k+2}+\frac{1}{(k+1)(k+2)}

Combine the fractions:

\frac{k(k+2)+1}{(k+1)(k+2)}

Distribute:

\frac{k^2+2k+1}{(k+1)(k+2)}

Factor the numerator:

\frac{(k+1)^2}{(k+1)(k+2)}

Cancel a common factor of (k+1)

\frac{k+1}{k+2}

We have proven the given expression and our formula for the sum are equal for all natural numbers,n.

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Is that a dice your talking about?
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3 years ago
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