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n200080 [17]
2 years ago
7

24 students in a class took an algebra test.

Mathematics
2 answers:
Vlada [557]2 years ago
8 0

Answer:

75 % passed

25 % did NOT pass

Step-by-step explanation:

vodomira [7]2 years ago
8 0
75% passed
25% did not pass
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Which number is composite A 19 B 23 C 33 D 37​
raketka [301]

Option C is correct.

Step-by-step explanation:

We need to identify the composite numbers from the given options

<u>Composite Numbers</u>

A number is composite number if it has factors other than 1 and the number itself i.e the number is not prime.

The numbers 19, 23 and 37 are prime numbers because they have factors 1 and the number itself while 33 is composite number because it has factors 1,3,11,33.

So, Option C is correct.

Keywords: Composite Numbers

Learn more about Composite Numbers at:

  • brainly.com/question/10941043
  • brainly.com/question/4706270

#learnwithBrainly

7 0
2 years ago
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Locate the point (-3.5, -1) on the coordinate plane:
MrRissso [65]

Answer:

go to the left 3.5 units then go down one and that's your point

8 0
2 years ago
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PLEASE HELP making brainliest if correct!!
dusya [7]

Answer:

A- sss

Step-by-step explanation:

5 0
2 years ago
Prove or disprove (from i=0 to n) sum([2i]^4) &lt;= (4n)^4. If true use induction, else give the smallest value of n that it doe
ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for n\geq 78

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: \sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4

For n=1: \sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4

From this point we will assume that n\geq 2

As we can see, \sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4 and (4n)^4=256n^4. Then,

\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4

Now, we will use the formula for the sum of the first 4th powers:

\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}

Therefore:

\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0

and, because n \geq 0,

465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1

Observe that, because n \geq 2 and is an integer,

n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5

In concusion, the statement is true if and only if n is a non negative integer such that n\leq 77

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute  (4n)^4- \sum^{n}_{i=0} (2i)^4 for 77 and 78 you will obtain:

(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064

(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992

7 0
3 years ago
Explain how to find 27% of 16 using multiplication by
just olya [345]

Answer:

4.32

Step-by-step explanation:

You can multiply 16 by 0.27 to get the answer.

16*0.27=4.32

You can check your work with estimation because 27% is about 1/4 of the number, and just by looking at the number, 4 is 1/4 of the number.

6 0
2 years ago
Read 2 more answers
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