Need help PLEASE HELP ME WITH MY HMK

The quality-control department of Starr Communications, a manufacturer of video-game DVDs, has determined from records that 1.5%

Sunny_sXe [5.5K]

Let A represents the event that DVD has a video effect.
So,
P(A) = 1.5 %

Let B represents the event that DVD has a audio defect.
So,
P(B) = 0.8 %

A ∩ B represents that DVD has both audio and video defect.
So,
P(A∩B) = 0.4%

Part a)
A ∪ B represents that DVD has a video or audio defect.

P(A∪B) = P(A)+P(B)-P(A∩B) = 1.5 + 0.8 - 0.4 = 1.9%

Thus the probability that the DVD purchased by the customer has a video or audio defect is 1.9%.

Part b)
The complement of P(A∩B) will represent the probability that the DVD does not have video or audio defect.

So, the probability that DVD purchased by customer does not have video or audio defect is 1 - 0.4 = 0.96%

3 0

3 years ago

PLZZZ ANSWER QUICK!!!

GREYUIT [131]

Answer:

A. x - 38.45 = 17.71

Step-by-step explanation:

The price of the shirt subtracted off the amount of money she had before would give 17.71

7 0

3 years ago

F(x)=3x-6<br> how do I figure this out

vitfil [10]

F(x) = 3x-6
add 6 to each side
6=3x
dived each side by 3
2=x

3 0

4 years ago

Read 2 more answers

Need help will give branliest also how u got answer

erastova [34]

Answer:

x= -3

Step-by-step explanation:

simplify.

1+4=x-8

Move all terms containing x to the left

1+3x=-8

Move all terms NOT containing to the right

3x=-9

Divide each term by 3

x= -3

7 0

3 years ago

Read 2 more answers

The countries of Europe report that 46% of the labor force is female. The United Nations wonders if the percentage of females in

Anestetic [448]

Answer:

a) Point estimate p=0.43.

b) The 95% confidence interval for the population proportion of female workers in US is (0.388, 0.473).

c) This confidence interval tells us that, with 95% confidence, the true proportion of females in the U.S. labor force is within 0.388 and 0.473.

d) We can conclude that females do not make up a majority of the US labor force, because we can claim, with 95% confidence, that the true proportion of female workers is under 0.473, which is below 0.5 (or 50%).

Step-by-step explanation:

a) A point estimate can be calculated using the sample results and dividing the amount of female workers in the sample by the sample size:

$p=X/n=226/525=0.43$

b) We have to calculate a 95% confidence interval for the proportion.

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.43*0.57}{525}}\\\\\\ \sigma_p=\sqrt{0.000467}=0.022$

The critical z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=1.96 \cdot 0.022=0.042$

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z \cdot \sigma_p = 0.4305-0.042=0.388\\\\UL=p+z \cdot \sigma_p = 0.4305+0.042=0.473$

The 95% confidence interval for the population proportion of female workers in US is (0.388, 0.473).

c) This confidence interval tells us that, with 95% confidence, the true proportion of females in the U.S. labor force is within 0.388 and 0.473.

d) We can conclude that females do not make up a majority of the US labor force, because we can claim, with 95% confidence, that the true proportion of female workers is under 0.473, which is below the 50%.

If the confidence interval have included the value 0.5, we could not be sure (with 95% confidence) that female workers do not represent de majority, as there would be possibility that the true proportion is 0.5.

3 0

4 years ago