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Sunny_sXe [5.5K]
4 years ago
5

The value of √30 is between which two to numbers ?

Mathematics
1 answer:
Naddika [18.5K]4 years ago
8 0
5 and 6, the value is about 5.48
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How to solve the problem
Dima020 [189]
A=1 B=2 C=3 Yes it would work for all because its using the commutative property.
5 0
3 years ago
Can someone please help?
Pepsi [2]

Answer:

Try working from Right to left

Step-by-step explanation:

That way the exponits will work out.

6 0
4 years ago
A number picked at random from the numbers 1 through 15 is prime. This "question" is on my study guide for Unit 6 Probability an
Sphinxa [80]

Solution:

we are given that

A number picked at random from the numbers 1 through 15 is prime.

As we know the number between 1 and 15 which are primes are listed below

2,3,5,7,11,13.

Here total number of favourable outcomes are 6.

The numbers 1 through 15 are as below:

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15.

Here total number of possible outcoes are 15.

Hence the probability that if a number picked at random from the numbers 1 through 15 is prime=\frac{6}{15}=\frac{2}{5}=0.40

Hence the probability that if a number picked at random from the numbers 1 through 15 is prime is 0.4

6 0
4 years ago
Read 2 more answers
Is 6 to the 3rd power equal to 2•2•2•3•3•3? Full explaination and reasoning.
Zolol [24]

Answer:

yes

Step-by-step explanation:

if you do 6^3 you get 216

6*6*6=216

6*6=36

36*6=216

2*2*2*3*3*3=216

2*2=4

4*2=8

8*3=24

24*3=72

72*3=216

4 0
4 years ago
What is the simplified form of the expression k^3(k7/5)^-5
saveliy_v [14]
The expression can be simplified as:

k^3(k7/5)^-5
= k^(3+-5) * (7/5)^-5
(Collecting the powers of k at one side and the constants at other side)
= k^-2 * (5/7)^5
(Solving thr integer powers)
= k^-2 * (3125/16807)
4 0
3 years ago
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