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iren [92.7K]
3 years ago
14

Number line with open circle on 9 and shading to the left. Which of the following inequalities best represents the graph above?

x < 9 x > 9 x ≤ 9 x ≥ 9
Mathematics
1 answer:
earnstyle [38]3 years ago
3 0
X is less than 9. We know it is only less than and not equal to because the dot is open. Just as well, because the line is shaded to the left of this point, we know x's value is decreasing; thus making it less than 9.

I hope this Helps!
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Estimate √97.5 / 1.96
evablogger [386]

Answer: 5


Step-by-step explanation:

 1. You have the following expression given in the problem above:

\frac{\sqrt{97.5}}{1.96}

2. You can estimate the result by rounding the numerator and the denominator.

3. As you can see, you can round up 97.5 to 100.

4. Then, you can round up 1.96 to 2.

5. Therefore, you have:

\frac{\sqrt{100}}{2}

\frac{10}{2}=5

6. Therefore, the result is 5.

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3 years ago
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Oksi-84 [34.3K]
8 times 3 = 24 +4= 28

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5 0
3 years ago
Read 2 more answers
4 times as much as 3 is blank
NeX [460]

Answer:

12

Step-by-step explanation:

4 times 3 is 12.

4+4+4=12

5 0
3 years ago
9. A large electronic office product contains 2000 electronic components. Assume that the probability that each component operat
Marysya12 [62]

Answer:

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it works correctly, or it does not. The probability of a component falling is independent from other components. 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.

In this problem we have that:

n = 2000, p = 1-0.995 = 0.005

Approximate the probability that five or more of the original 2000 components fail during the useful life of the product.

We know that either less than five compoenents fail, or at least five do. The sum of the probabilities of these events is decimal 1. So

P(X < 5) + P(X \geq 5) = 1

We want P(X \geq 5)

So

P(X \geq 5) = 1 - P(X < 5)

In which

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

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

P(X = 0) = C_{2000,0}.(0.005)^{0}.(0.995)^{2000} = 0.000044

P(X = 1) = C_{2000,1}.(0.005)^{1}.(0.995)^{1999} = 0.000445

P(X = 2) = C_{2000,2}.(0.005)^{2}.(0.995)^{1998} = 0.002235

P(X = 3) = C_{2000,3}.(0.005)^{3}.(0.995)^{1997} = 0.007480

P(X = 4) = C_{2000,4}.(0.005)^{4}.(0.995)^{1996} = 0.018765

P(X < 5) = P(X = 0) + `P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.000044 + 0.000445 + 0.002235 + 0.007480 + 0.018765 = 0.0290

P(X \geq 5) = 1 - P(X < 5) = 1 - 0.0290 = 0.9710

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.

4 0
3 years ago
If the original length are multiple by 1/3 what are the new coordinates
Olenka [21]

Answer:

2

Step-by-step explanation:

7 0
4 years ago
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