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ella [17]
3 years ago
9

Move the values to the boxes to complete the inequality statement.

Mathematics
1 answer:
Licemer1 [7]3 years ago
5 0

Answers:

  • first blank =  not sure (see explanation below)
  • second blank = 3*sqrt(6)
  • third blank = 3pi

============================================================

Explanation:

Use your calculator to compute 3*sqrt(6) = 7.348469 which is approximate.

We can see that 3*sqrt(6), aka 7.348469, is between 7 and 9

Do the same for 3pi to get 3pi = 9.424778 approximately. So this goes between 9 and 11

As for the first blank, between 5 and 7, it seems like an answer choice is missing. So I would ask your teacher about that. If your teacher meant it to be left blank, then just ignore it.

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Simora [160]
M >= 85. The >= sign means greater than or equal to.
8 0
3 years ago
Expanding polynomial functions using binomial theorem or pascals triangle
konstantin123 [22]

Answer:

The formula for Pascal's Triangle comes from a relationship that you yourself might be able to see in the coefficients below.

(x + y)0 (x + y)1 (x + y)² (x + y)3 (x + y)41 x + y x² + 2xy + y² x3 + 3x2Y + 3xY2 + y3 x4 + 4x3Y + 6x2Y2 + 4XY3 + Y4

HOPE THIS HELPS!

7 0
3 years ago
Ivanna needs to buy some pencils. Brand A has a pack of 36 pencils for 8.52 . Brand B has a pack of 48 pencils for 9.98 . Find t
Taya2010 [7]

Answer:

Brand B is better

Step-by-step explanation:

Brand A's unit price is $0.24

Brand B's unit price is $0.21

5 0
2 years ago
Refer to the Trowbridge Manufacturing example in Problem 2-35. The quality control inspection proce- dure is to select 6 items,
Ivanshal [37]

Answer:

77.64% probability that there will be 0 or 1 defects in a sample of 6.

Step-by-step explanation:

For each item, there are only two possible outcomes. Either it is defective, or it is not. The probability of an item being defective is independent of other items. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The true proportion of defects is 0.15

This means that p = 0.15

Sample of 6:

This means that n = 6

What is the probability that there will be 0 or 1 defects in a sample of 6?

P(X \leq 1) = P(X = 0) + P(X = 1)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{6,0}.(0.15)^{0}.(0.85)^{6} = 0.3771

P(X = 1) = C_{6,1}.(0.15)^{1}.(0.85)^{5} = 0.3993

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.3771 + 0.3993 = 0.7764

77.64% probability that there will be 0 or 1 defects in a sample of 6.

5 0
3 years ago
How to solve this problem
ANTONII [103]

Answer:

B: (2, -1)

Step-by-step explanation:

1) First isolate the y in both equations

2) Set the equations equal to each other

3) Solve for x (you should get 2 and 5)

4) Insert the x values back in to get your y values

5) You should have gotten (2, -1) and (5, 2)

These are your two answers, but the question is only asking for one solution and (5,2) isn't one of the options, so it has to be (2,-1).

8 0
3 years ago
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