A second grade student is 4 feet tall. Her teacher is 5 2/3 feet tall. How many times as tall as the student is the teacher

eimsori [14]

Use the FOIL method (First, Outside, Inside, Last)

6r(-8r) = -48r²
6r(-3) = -18r
-1(-8r) = 8r (note: two negatives multiplied together = positive answer)
-1(-3) = 3

-48r² - 18r + 8r + 3

Combine like terms:

-48r² - 18r + 8r + 3

-48r² - 10r + 3

-48r² - 10r + 3 is your answer

hope this helps

6 0

2 years ago

Pedro has created the function f(x)= 4x-3/2 to represent the number of assingments he has completed where x represents the numbe

Law Incorporation [45]

The given function is
f(x) = 4x - 3/2
where
f(x) = number of assignments completed
x = number of weeks required to complete the assignments

We want to find f⁻¹ (30) as an estimate of the number of weeks required to complete 30 assignments.
The procedure is as follows:

1. Set y = f(x)
y = 4x - 3/2

2. Exchange x and y
x = 4y - 3/2

3. Solve for y
4y = x + 3/2
y = (x +3/2)/4

4. Set y equal to f⁻¹ (x)
f⁻¹ (x) = (x + 3/2)/4

5. Find f⁻¹ (30)
f⁻¹ (30) = (30 + 3/2)/4 = 63/8 = 8 (approxmately)

Answer:
Pedro needs about 8 weeks to complete 30 assignments.

6 0

1 year 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

The number of hours Layla worked each week for the first eight weeks of summer break are recorded below.

Rainbow [258]

Answer: The value of the center increases and the distribution has a smaller spread.

Step-by-step explanation:

Re - arranging the first value in ascending order , we have :

12 ,14, 16 , 19, 21 , 22 , 28 , 32

Median = 19 + 21 / 2

= 40/2

= 20

Also calculating the standard variation in order to know the spread , we need to first of all calculate the mean

Mean = 12 + 14 + 16 + 19 + 21 + 22 + 28 + 32 / 8

Mean = 164/8

Mean = 20.5

Therefore to calculate the standard deviation, we need to calculate the variance, which is

$(12-20.5)^{2}$ + $(14-20.5)^{2}$ + $(16-20.5)^{2}$ + $(19-20.5)^{2}$ + $(21-20.5)^{2}$ + $(22-20.5)^{2}$ + $(28-20.5)^{2}$ + $(32-20.5)^{2}$ / 8

Variance = 328/8

Variance = 41

Standard deviation = $\sqrt{variance}$

S.D = $\sqrt{41}$

S.D = 6.4

Also re arranging the second values , we have :

20 , 24 , 25 , 31

Median = 24 + 25 / 2

Median = 49/2

Median = 24.5

Calculating the S.D using the same format , the S.D = 3.9

Comparing the two we can conclude that the value at the center increases and the distribution has a smaller spread

8 0

2 years ago

Read 2 more answers

On a december day, the probability of snow is .30. the probability of a "cold" day is .50. the probability of snow and "cold" we

Svetradugi [14.3K]

Events A and B are independent if the following holds true:

P(A ∩ B) = P(A) * P(B)

where P(A ∩ B) is the probability of A and B, P(A) is the probability of A, and P(B) is the probability of B.

Setting A = "snow" and B = "cold weather", and plugging in the above numbers will give you the answer.

7 0

2 years ago