Will qamar have enough lacquer to cover the floor below? Please help I have been stuck on this for three days!!!

The price of a pair of headphones is $39.99 before tax. Dalton bought the headphones for 30% off. He then paid 6% sales tax on t

dimaraw [331]

Answer:

part A 1.68

part B 29.67

Step-by-step explanation:

First you multiply $39.99 dollars and 0.30 = $11.9970 you round this to $12.00

After you apply the discount $39.99 - $12.00 = 27.99

Next you must multiply 27.99*0.06 =1.6794 and round to 1.68 to get tax

Last you add $27.99 and 1.68 to get a total of $29.67 for the headphones

Plz give brainliest i didnt look it up.

4 0

2 years ago

Before any dinner service, the manager confirms when the number of reservations at a restaurant is subtracted from twice the num

pochemuha

Answer:

Its A

Step-by-step explanation:

4 0

2 years ago

A certain article indicates that in a sample of 1,000 dog owners, 610 said that they take more pictures of their dog than of the

lbvjy [14]

Answer:

(a) The 90% confidence interval is: (0.60, 0.63) Correct interpretation is (3).

(b) The 95% confidence interval is: (0.42, 0.46) Correct interpretation is (1).

(c) First, the confidence level in part(b) is more than the confidence level in part(a) is, so the critical value of part(b) is more than the critical value of part(a), Second the margin of error in part(b) is more than than in part(a).

Step-by-step explanation:

(a)

Let X = number of dog owners who take more pictures of their dog than of their significant others or friends.

Given:

X = 610

n = 1000

Confidence level = 90%

The (1 - α)% confidence interval for population proportion is:

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

The sample proportion is:

$\hat p=\frac{X}{n}=\frac{610}{1000}=0.61$

The critical value of z for a 90% confidence level is:

$z_{\alpha/2}=z_{0.10/2}=z_[0.05}=1.645$

Construct a 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends as follows:

$CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.61\pm 1.645\sqrt{\frac{0.61(1-0.61}{1000}}\\=0.61\pm 0.015\\=(0.595, 0.625)\\\approx(0.60, 0.63)$

Thus, the 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends is (0.60, 0.63).

Interpretation:

There is a 90% chance that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends falls within the interval (0.60, 0.63).

Correct option is (3).

(b)

Let X = number of dog owners who are more likely to complain to their dog than to a friend.

Given:

X = 440

n = 1000

Confidence level = 95%

The (1 - α)% confidence interval for population proportion is:

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

The sample proportion is:

$\hat p=\frac{X}{n}=\frac{440}{1000}=0.44$

The critical value of z for a 95% confidence level is:

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

Construct a 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend as follows:

$CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.44\pm 1.96\sqrt{\frac{0.44(1-0.44}{1000}}\\=0.44\pm 0.016\\=(0.424, 0.456)\\\approx(0.42, 0.46)$

Thus, the 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend is (0.42, 0.46).

Interpretation:

There is a 95% chance that the true proportion of dog owners who are more likely to complain to their dog than to a friend falls within the interval (0.42, 0.46).

Correct option is (1).

(c)

The confidence interval in part (b) is wider than the confidence interval in part (a).

The width of the interval is affected by:

The confidence level
Sample size
Standard deviation.

The confidence level in part (b) is more than that in part (a).

Because of this the critical value of z in part (b) is more than that in part (a).

Also the margin of error in part (b) is 0.016 which is more than the margin of error in part (a), 0.015.

First, the confidence level in part(b) is more than the confidence level in part(a) is, so the critical value of part(b) is more than the critical value of part(a), Second the margin of error in part(b) is more than than in part(a).

4 0

1 year ago

.

crimeas [40]

Answer:

coffee is the best answer in my mind

4 0

2 years ago

Leslie started last week with $1200 in her checking account. During the week, she wrote the checks below. Trans typ./ Check no.

Greeley [361]

Answer:d

Step-by-step explanation:

6 0

2 years ago