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3 years ago

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

1 answer:

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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To be clear i don't want the flat out answer i want help on how to solve it .

liberstina [14]

The answer to the problem is as follows:

So each thing separate by a positive, negative, multiplication or division sign. so here it would be 8h is one term and 15t is the second term so there are two.

I hope my answer has come to your help. God bless and have a nice day ahead!

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 X = 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 - α)% confidence interval for population proportion p is:

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

The critical value of z for 95% confidence level is,

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

*Use a z-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

Five years ago age of father was thrice the age of son. if son is 30 years old now what is current age of father...????

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

7 0

3 years ago