Add the following weights: 13 lb 7 oz + 10 lb 2 oz Ib OZ

how many liters of 40% oil dressing should be mixed with 4 liters of a dressing that is 25% oil to get a dressing that is 30% oi

Ronch [10]

Volume of first oil = A, volume of second oil = B, volume of mixture = M
A = ?, B = 4l, M = A + B

40%A + 25%B = 30%M
40%A + 25%4l = 30%(A+4l)
40%A + 25%4l = 30%A + 30%4l
40A + 100l = 30A + 120l
40A - 30 A = 120l - 100l
10A = 20l
A = 2l

Answer: 2 litres of 40% oil dressing.

Also note: That's an approximation, because the volumes are not strictly additive. For example: mixing 50ml pure ethanol with 50ml water will give you about 95ml of mixture. To get an accurate answer, you'd have to measure the volume of the final mixture and then divide your total oil content by that.

4 0

3 years ago

10q - 30 is equivalent

gladu [14]

Answer:

3

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

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

2 years ago

Some triangles can have more than one obtuse angle. True or false?

Reptile [31]

False. A triangle can't have more than one obtuse angle.

7 0

3 years ago

The m∠P is three less than twice the measure of ∠Q. If ∠P and ∠Q are supplementary angles, find the measures of both angles.

notka56 [123]

Answer:

m∠P =119

m∠Q = 61

Step-by-step explanation:

m∠P = 2(∠Q)-3 or p=2(q)-3

supplementary angles so they combine to equal 180!

soooooo we can take

q+p=180

input what p equals and you get

q+ 2(q)-3=180

3q-3=180

3q=183

q=61

now we have what m∠Q is, so just subtract that from 180 to get m∠P

180-61=119

7 0

3 years ago

Add the following weights:13 lb 7 oz+ 10 lb 2 ozIbOZ​

Add the following weights:13 lb 7 oz+ 10 lb 2 ozIbOZ