There are 150 kangaroos at a wildlife park there are 66 female what percentage is female

Two groups of students were asked how many pets they had. The table below shows the numbers for each group:

Nana76 [90]

where are the answers?

5 0

2 years ago

Read 2 more answers

Blue's Berry Farm charges Percy a total of

kaheart [24]

Answer:

The price for each kilogram of strawberries is $7.50

Step-by-step explanation:

The question is

Blues berry farm charges Percy a total of $24.75 for entrance and 2.5 kilograms of strawberries. The entrance fee is $6 and the price for each kilogram of strawberries is constant

Determine the price for each kilogram of strawberries

Let

x ----> the price for each kilogram of strawberries

we know that

The entrance fee plus the price of each kilogram of strawberries multiplied by the number of kilogram of strawberries must be equal to $24.75

so

The linear equation that represent this situation is

$2.5x+6=24.75$

solve for x

subtract 6 both sides

$2.5x=24.75-6\\2.5x=18.75$

divide by 2.5 both sides

$x=\$7.50$

therefore

The price for each kilogram of strawberries is $7.50

8 0

2 years ago

Which expressions can you use to find 75% of 56? Check all that apply.

alisha [4.7K]

Your answer would be B and C

7 0

2 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

A list has 10 numbers. Each number is a 1, 2, or 3. The average is 2 and the SD is 0. What is the list?

OLga [1]

Answer:

1, 1, 1, 2, 2, 2, 2, 3, 3, 3

Step-by-step explanation:

Since the standard deviation = 0

And the mean of the number = 2

If X is the sum of all the numbers, then

X/10 = 2

Cross multiply and make X the subject of formula

X = 2 × 10 = 20

The sum of the numbers will be 20

The number will be

1, 1, 1, 2, 2, 2, 2, 3, 3, 3

The sum of the above numbers is equal to 20 and the average value of the numbers is equal to 2.

7 0

2 years ago