Choose the expression that is equivalent to the expression

In a set of 25 aluminum castings, four castings are defective (D), and the remaining twenty-one are good (G). A quality control

lianna [129]

Answer:

The sample space for selecting the group to test contains 2,300 elementary events.

Step-by-step explanation:

There are a total of N = 25 aluminum castings.

Of these 25 aluminum castings, n₁ = 4 castings are defective (D) and n₂ = 21 are good (G).

It is provided that a quality control inspector randomly selects three of the twenty-five castings without replacement to test.

In mathematics, the procedure to select k items from n distinct items, without replacement, is known as combinations.

The formula to compute the combinations of k items from n is given by the formula:

${n\choose k}=\frac{n!}{k!(n-k)!}$

Compute the number of samples that are possible as follows:

${25\choose 3}=\frac{25!}{3!\times (25-3)!}$

$=\frac{25\times 24\times 23\times 22!}{3!\times 22!}\\\\=\frac{25\times 24\times 23}{3\times 2\times 1}\\\\=2300$

The sample space for selecting the group to test contains 2,300 elementary events.

6 0

3 years ago

The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

a

$336.04 < \mu < 443.96$

b

The margin of error will increase

c

The margin of error will decreases

d

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

So

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

$n \ \ \alpha \ \ \frac{1}{E^2 }$

So if the sample size increases the margin of error decrease

The sample proportion is mathematically represented as

$\r p = \frac{210}{500}$

$\r p = 0.42$

Given that the confidence level is 0.99 the level of significance is $\alpha = 0.01$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} }* \sqrt{ \frac{\r p (1- \r p )}{n} }$

=> $E = 0.42 * \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }$

=> $E = 0.0093$

The 99% confidence interval is

$\r p - E < p < \r p + E$

$0.42 - 0.0093 < p < 0.42 + 0.0093$

$0.4107 < p < 0.4293$

4 0

3 years ago

I need to know what is 18f=-216

mamaluj [8]

Divide 18 to both sides and you get -12

8 0

3 years ago

Solve for x. Enter the solutions from least to greatest. Round to two decimal places. (x + 8)^2 - 7 = 0

Karolina [17]

Answer:

x

= − 5.35424868 , − 10.64575131

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

The length of a rectangle is 12 in. and the perimeter is 56 in. find the width of the rectangle

pentagon [3]

The perimeter p of the rectangle with length l and width w is:
p = 2l + 2w
that is, two times its length and two times its width, lets solve for the width w and substitute known data:
p = 2l + 2w
2w = p - 2l
w = (p - 2l)/2
w = (56 - (2*12))/2
w = (56 - 24)/2
w = 32/2
w = 16
therefore the width of the rectangle is 16 in

8 0

3 years ago