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Digiron [165]
2 years ago
5

40 students at a particular school are randomly selected for a survey. 18 have blue eyes. If there are 400 students at this scho

ol, what is a reasonable estimate of the total number of students with blue eyes?
Mathematics
2 answers:
podryga [215]2 years ago
7 0
A reasonable estimate would be 180 students with blue eyes
barxatty [35]2 years ago
3 0

Answer:

180 students have blue eyes

Step-by-step explanation:

If 18/40 students have blue eyes, when the total number rof students is 400

18/40 × 400=180

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Fudgin [204]

Answer:

A) 13

Step-by-step explanation:

If you recall the formula for finding a diagonal, or a^2 + B^2 = C^2, you will get 12^2 + 5^2 = C^2, if you solve that, you will get 13!

4 0
2 years ago
Use elimination method to solve<br> 5x+y=-10<br> 7x-3y=-36
k0ka [10]
5x+y=-10\\ \\ 7x-3y=-36

Let's multiply the first equation by 3. (As you can see y's coefficient in the first one is 1 and in the second one is -3 , we will multiply the first equation by 3 so when we add the equations their sum will be 0)

3\cdot (5x+y)=3\cdot -10\\ \\ 15x+3y=-30

Now let's add this new equation and our second equation.

7x-3y=-36\\ \\ 15x+3y=-30\\ \\ (15x+3y)+(7x-3y)=(-30)+(-36)\\ \\ 15x+7x+3y-3y=-30-36\\ \\ 22x=-66\\ \\ x=\frac { -66 }{ 22 } \\ \\ x=-3

We found x=-3

Now let's plug x's value in one of the equations to find y's value.

x=-3\\ \\ 5x+y=-10\\ \\ 5\cdot -3+y=-10\\ \\ -15+y=-10\\ \\ y=-10+15\\ \\ y=5

So we found y=5

Solution ;

(-3, 5)

7 0
3 years ago
Help plz:)))I’ll mark u Brainliest
MArishka [77]
This a right triangle shown above
8 0
3 years ago
Read 2 more answers
A small combination lock on a suitcase has 55 ​wheels, each labeled with the 10 digits 0 to 9. How many 55 digit combinations ar
den301095 [7]
Suppose a_n is the number of possible combinations for a suitcase with a lock consisting of n wheels. If you added one more wheel onto the lock, there would only be 9 allowed possible digits you can use for the new wheel. This means the number of possible combinations for n+1 wheels, or a_{n+1} is given recursively by the formula

a_{n+1}=9a_n

starting with a_1=10 (because you can start the combination with any one of the ten available digits 0 through 9).

For example, if the combination for a 3-wheel lock is 282, then a 4-wheel lock can be any one of 2820, 2821, 2823, ..., 2829 (nine possibilities depending on the second-to-last digit).

By substitution, you have

a_{n+1}=9a_n=9^2a_{n-1}=9^3a_{n-2}=\cdots=9^na_1=10\times9^n

This means a lock with 55 wheels will have

a_{55}=10\times9^{54}

possible combinations (a number with 53 digits).
7 0
3 years ago
Jan has $245.50 to spend on home repairs. The repairman charges $75 for the initial visit and $35 per hour or any part of an hou
Murrr4er [49]

Answer:

4

Step-by-step explanation:

4 0
3 years ago
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