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NNADVOKAT [17]
3 years ago
11

What type of sequence is the following?

Mathematics
1 answer:
zubka84 [21]3 years ago
4 0
I think the sequence is: 1001.02, 1001.06, 1001.1, 1001.14

As you can observe, the next number is the sum of the previous number plus 0.04.

The above sequence is an Arithmetic sequence. The movement from one term to the other is by adding or subtracting a constant number.


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Solve the equation 2(x-10)= 24 , what is the value of x
N76 [4]

Answer:

x=22

Step-by-step explanation:

2(x-10)=24

Distribute the parenthesis

2x-20=24

Add 20 to both sides

2x=44

Divide both sides by 2

\frac{2x}{2} =\frac{44}{2}

x=22

Hope this helps

5 0
3 years ago
Read 2 more answers
What is the square and square root of 9x⁸​
adelina 88 [10]

Answer:

square: (9x^8)^2= 81x^16

square root: (9x^8)^1/2 = 3x^4

Step-by-step explanation:

the ^ mean that the following num should be like this ​⁸​

8 0
3 years ago
What is the solution of this system of linear equations?
Allisa [31]

Answer:

(O,2)

Step-by-step explanation:

So I signed in to iReady did the question and got it right.

7 0
3 years ago
student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
Solve the following system to find the point(s) of intersection. <br> y = x² - 2x - 3<br> y = 2x - 3
omeli [17]

x=4 y=5

if we put two pharases together

the answer is as follows pic :

x=4 y=5

5 0
3 years ago
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