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valina [46]
3 years ago
5

Jack says 0.90 is between 0.9 and 0.91. Is he correct? Explain.

Mathematics
1 answer:
OleMash [197]3 years ago
3 0
Well, 0.90 is actually equal to 0.9. It is less than 0.91 and equal to 0.9 so that means it is not correct. 

I hope this helps:)
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liberstina [14]
The answer to the problem is as follows:

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4 0
3 years ago
A truck loaded with 8000 electronic circuit boards has just pulled into a firm’s receiving dock. The supplier claims that no mor
Juliette [100K]

Answer:

The 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

Step-by-step explanation:

Let <em>X</em> = number of boards that fall outside the most rigid level of industry performance specifications.

In a random sample of 300 boards the number of defective boards was 12.

Compute the sample proportion of defective boards as follows:

\hat p =\frac{12}{300}=0.04

The (1 - <em>α</em>)% confidence interval for population proportion <em>p</em> is:

CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

The critical value of <em>z</em> for 95% confidence level is,

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

*Use a <em>z</em>-table.

Compute the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification as follows:

CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.04\pm1.96\sqrt{\frac{0.04(1-0.04)}{300}}\\=0.04\pm0.022\\=(0.018, 0.062)\\\approx(1.8\%, 6.2\%)

Thus, the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

3 0
3 years ago
What is the easiest way to solve a quadratic equation?​
algol [13]

Answer: This is an opinion, look below! :)

Step-by-step explanation:

Factor the expression. To factor the expression, you have to use the factors of the {\displaystyle x^{2}}x^{2} term (3), and the factors of the constant term (-4), to make them multiply and then add up to the middle term, (-11). Here's how you do it:

Since {\displaystyle 3x^{2}}3x^{2} only has one set of possible factors, {\displaystyle 3x}3x and {\displaystyle x}x, you can write those in the parenthesis: {\displaystyle (3x\pm ?)(x\pm ?)=0}(3x\pm ?)(x\pm ?)=0.

Then, use process of elimination to plug in the factors of 4 to find a combination that produces -11x when multiplied. You can either use a combination of 4 and 1, or 2 and 2, since both of those numbers multiply to get 4. Just remember that one of the terms should be negative, since the term is -4.[3]

By trial and error, try out this combination of factors {\displaystyle (3x+1)(x-4)}(3x+1)(x-4). When you multiply them out, you get {\displaystyle 3x^{2}-12x+x-4}3x^{2}-12x+x-4. If you combine the terms {\displaystyle -12x}-12x and {\displaystyle x}x, you get {\displaystyle -11x}-11x, which is the middle term you were aiming for. You have just factored the quadratic equation.

As an example of trial and error, let's try checking a factoring combination for {\displaystyle 3x^{2}-11x-4=0}3x^{2}-11x-4=0 that is an error (does not work): {\displaystyle (3x-2)(x+2)}(3x-2)(x+2) = {\displaystyle 3x^{2}+6x-2x-4}3x^{2}+6x-2x-4. If you combine those terms, you get {\displaystyle 3x^{2}-4x-4}3x^{2}-4x-4. Though the factors -2 and 2 do multiply to make -4, the middle term does not work, because you needed to get {\displaystyle -11x}-11x, not {\displaystyle -4x}-4x.

7 0
3 years ago
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professor190 [17]

Answer:

The father is 75 years old

Step-by-step explanation:

30-5=25 years (sons age)

25×3=75 years old (father)

3 0
3 years ago
Evaluate all numerical expressions with exponents
Eduardwww [97]
Y to the power of 2 over 2x to the power of 3
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3 years ago
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