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bearhunter [10]
2 years ago
9

The dotplot below displays the number of math classes taken by a random sample of students at a high school.

Mathematics
1 answer:
s2008m [1.1K]2 years ago
5 0

Answer:

C-  The distribution of math classes is unimodal symmetric with a center around 4 classes.

Step-by-step explanation:

On edge 2022

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A space shuttle orbits Earth one time in 90 minutes how many times does it orbit around Earth in 6 hours
amid [387]
So their are 60 min in an hour. you multiply 60 min times 6 (hours). you get 360 min. Then you divide that by 90 min and you get 4
Answer it orbits 4 times
hope that helps

4 0
4 years ago
Read 2 more answers
What is 35% of 98???
Serga [27]

Answer:34.3

Step-by-step explanation:35% of 98 = 34.3

4 0
3 years ago
URGENT The number of counties in state A and the number of counties in state B are consecutive even integers whose sum is 158. I
Daniel [21]
I’m not sure try using photo math
5 0
3 years ago
If p q r is prime numbers such that pq+r=73, what is the least possible value of p+q+r
Jobisdone [24]
The answer to this question would be: p+q+r = 2 + 17 + 39= 58

In this question, p q r is a prime number. Most of the prime number is an odd number. If p q r all odd number, it wouldn't be possible to get 73 since
odd x odd + odd= odd + odd = even
Since 73 is an odd number, it is clear that one of the p q r needs to be an even number. 

There is only one odd prime number which is 2. If you put 2 in the r the result would be:
pq+2= 73
pq= 71
There will be no solution for pq since 71 is prime number. That mean 2 must be either p or q. Let say that 2 is p, then the equation would be: 2q + r= 73

The least possible value of p+q+r would be achieved by founding the highest q since its coefficient is 2 times r. Maximum q would be 73/2= 36.5 so you can try backward from that. Since q= 31, q=29, q=23 and q=19 wouldn't result in a prime number r, the least result would be q=17
r= 73-2q
r= 73- 2(17)
r= 73-34=39

p+q+r = 2 + 17 + 39= 58
6 0
3 years ago
Read 2 more answers
Factor completely.<br> 150m^2nz+20mn^2c-120m^2nc-25mn^2z
Anika [276]

Answer:

The factor form is:

5mn\left(6m-n\right)\left(5z-4c\right)

Step-by-step explanation:

Here we have to find the factor of the expression:

150m^2nz+20mn^2c-120m^2nc-25mn^2z

So we need to take out the common terms, as we do for finding the greatest common factors.

Now the expression that is given can be re-written as:

150m^2nz+20mn^2c-120m^2nc-25mn^2z\\=150mmnz+20mnnc-120mmnc-25mnnz\\=30\cdot \:5nmmz+4\cdot \:5nmnc-24\cdot \:5nmmc-5\cdot \:5nmnz

Next, we will find the common terms, as follows;

30\cdot \:5nmmz+4\cdot \:5nmnc-24\cdot \:5nmmc-5\cdot \:5nmnz\\=5nm\left(30mz+4nc-24mc-5nz\right)\\

Now  we will factorise the expression:

\left(30mz+4nc-24mc-5nz\right)\\

as follows;

\left(30mz+4nc-24mc-5nz\right)\\=6m\left(5z-4c\right)+n\left(4c-5z\right)\\=\left(-4c+5z\right)\left(6m-n\right)\\

So the final factor form is:

150m^2nz+20mn^2c-120m^2nc-25mn^2z=5mn\left(6m-n\right)\left(5z-4c\right)

8 0
3 years ago
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