Need help someone please

A quality control engineer is interested in estimating the proportion of defective items coming off a production line. In a samp

fenix001 [56]

Answer:

The lower bound of a 99% C.I for the proportion of defectives = 0.422

Step-by-step explanation:

From the given information:

The point estimate = sample proportion $\hat p$

$\hat p = \dfrac{x}{n}$

$\hat p = \dfrac{55}{100}$

$\hat p$ = 0.55

At Confidence interval of 99%, the level of significance = 1 - 0.99

= 0.01

$Z_{\alpha/2} =Z_{0.01/2} \\ \\ = Z_{0.005} = 2.576$

Then the margin of error $E = Z_{\alpha/2} \times \sqrt{\dfrac{\hat p(1-\hat p)}{n}}$

$E = 2.576 \times \sqrt{\dfrac{0.55(1-0.55)}{100}}$

$E = 2.576 \times \sqrt{\dfrac{0.2475}{100}}$

$E = 2.576 \times0.04975$

E = 0.128156

E ≅ 0.128

At 99% C.I for the population proportion p is: $\hat p - E$

= 0.55 - 0.128

= 0.422

Thus, the lower bound of a 99% C.I for the proportion of defectives = 0.422

6 0

2 years ago

535053424567 x 6566207654999=???

valentina_108 [34]

Answer:

3.5132719e+24

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Solve the inequality x-4 < - 3.Then graph the solution set.show your work.

bearhunter [10]

Answer:

x < 1

Step-by-step explanation:

x - 4 < - 3

<u> + 4 +4</u> Do inverse operations by adding 4 on both sides

x < 1

Therefore, the answer is x < 1.

(It isn't allowing me to attach a photo but I am just going to try make a visual of the graph below)

<------------------------------o------------------>

-4 -3 -2 -1 0 1 2 3 4

The circle is open because based on the concept that in inequalities, <em>less/greater than and/or equal to (such as these ≥ ≤ )</em> signs are graphed with closed circles, while less/greater than (like these > <) signs are graphed with open circles. The direction of the arrow is moving to the left on the number line because all possible values for x would be less than 1. When graphing the inequality you would use a solid line rather than a dashed line (I had to improvise since I could not attach an image).

8 0

2 years ago

Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .

kogti [31]

Answer:

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if $(x+y)^{2} = x^{2} + 2\cdot x\cdot y + y^{2}$ . From statement, we must fulfill the following identity:

$a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}$

By Associative and Commutative properties, we can reorganize the expression as follows:

$a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2}$ (1)

Then, we have the following system of equations:

$x = a$ (2)

$(b^{2}-1) = 2\cdot x\cdot y$ (3)

$y = c$ (4)

By (2) and (4) in (3), we have the following expression:

$(b^{2} - 1) = 2\cdot a \cdot c$

$b^{2} = 1 + 2\cdot a \cdot c$

$b = \sqrt{1 + 2\cdot a\cdot c}$

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, $a, b, c > 1$ . If $a$ , $b$ and $c$ are prime numbers, then $2\cdot a\cdot c$ must be an even composite number, which means that $a$ and $c$ can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.