Which inequality is true A B C D Check picture

To multiply a monomial by a polynomial, use the (Associative or Distributive) Property. Using this property, the product of (2k^

JulijaS [17]

I'd do this in reverse order. Mult. each of the 3 terms inside parentheses by 4:

8k^2 - 28k + 12

6 0

3 years ago

Approximate the area between the xxx-axis and h(x) = \dfrac{1}{7-x}h(x)= 7−x 1 h, (, x, ), equals, start fraction, 1, divided

tiny-mole [99]

Answer:

$\dfrac{47}{60}$ sq. units.

Step-by-step explanation:

The given function is

$h(x)=\dfrac{1}{7-x}$

We need to find the area between x-axis and the given function from x=2 to x=5.

Left Riemann sum formula of area:

$Area=\sum_{n=0}^{N-1}f(x_n)(\Delta x_n)$

For given question,

$Area=\sum_{n=2}^{5-1}f(x_n)(\Delta x_n)$

$Area=\sum_{n=2}^{4}f(x_n)(\Delta x_n)$

$Area=f(x_2)(3-2)+f(x_3)(4-3)+f(x_4)(5-4)$

Now,

$Area=\dfrac{1}{7-2}\times (1)+\dfrac{1}{7-3}\times (1)+\dfrac{1}{7-4}\times (1)$

$Area=\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{3}$

$Area=\dfrac{12+15+20}{60}$

$Area=\dfrac{47}{60}$

Therefore, the required area is $\dfrac{47}{60}$ sq. units.

5 0

3 years ago

In Case Study 19.1, we learned that about 56% of American adults actually voted in the presidential election of 1992, whereas ab

Radda [10]

Answer:

a) Confidence interval for 68% confidence level

= (0.548, 0.572)

Confidence interval for 95% confidence level

= (0.536, 0.584)

Confidence interval for 99.99% confidence level = (0.523, 0.598)

b) The sample proportion of 0.61 is unusual as falls outside all of the range of intervals where the sample mean can found for all 3 confidence levels examined.

c) Standardized score for the reported percentage using a sample size of 400 = 2.02

Since, most of the variables in a normal distribution should fall within 2 standard deviations of the mean, a sample mean that corresponds to standard deviation of 2.02 from the population mean makes it seem very plausible that the people that participated in this sample weren't telling the truth. At least, the mathematics and myself, do not believe that they were telling the truth.

Step-by-step explanation:

The mean of this sample distribution is

Mean = μₓ = np = 0.61 × 1600 = 976

But the sample mean according to the population mean should have been

Sample mean = population mean = nP

= 0.56 × 1600 = 896.

To find the interval of values where the sample proportion should fall 68%, 95%, and almost all of the time, we obtain confidence interval for those confidence levels. Because, that's basically what the definition of confidence interval is; an interval where the true value can be obtained to a certain level.of confidence.

We will be doing the calculations in sample proportions,

We will find the confidence interval for confidence level of 68%, 95% and almost all of the time (99.7%).

Basically the empirical rule of 68-95-99.7 for standard deviations 1, 2 and 3 from the mean.

Confidence interval = (Sample mean) ± (Margin of error)

Sample Mean = population mean = 0.56

Margin of Error = (critical value) × (standard deviation of the distribution of sample means)

Standard deviation of the distribution of sample means = √[p(1-p)/n] = √[(0.56×0.44)/1600] = 0.0124

Critical value for 68% confidence interval

= 0.999 (from the z-tables)

Critical value for 95% confidence interval

= 1.960 (also from the z-tables)

Critical values for the 99.7% confidence interval = 3.000 (also from the z-tables)

Confidence interval for 68% confidence level

= 0.56 ± (0.999 × 0.0124)

= 0.56 ± 0.0124

= (0.5476, 0.5724)

Confidence interval for 95% confidence level

= 0.56 ± (1.960 × 0.0124)

= 0.56 ± 0.0243

= (0.5357, 0.5843)

Confidence interval for 99.7% confidence level

= 0.56 ± (3.000 × 0.0124)

= 0.56 ± 0.0372

= (0.5228, 0.5972)

b) Based on the obtained intervals for the range of intervals that can contain the sample mean for 3 different confidence levels, the sample proportion of 0.61 is unusual as it falls outside of all the range of intervals where the sample mean can found for all 3 confidence levels examined.

c) Now suppose that the sample had been of only 400 people. Compute a standardized score to correspond to the reported percentage of 61%. Comment on whether or not you believe that people in the sample could all have been telling the truth, based on your result.

The new standard deviation of the distribution of sample means for a sample size of 400

√[p(1-p)/n] = √[(0.56×0.44)/400] = 0.0248

The standardized score for any is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (0.61 - 0.56)/0.0248 = 2.02

Standardized score for the reported percentage using a sample size of 400 = 2.02

Since, most of the variables in a normal distribution should fall within 2 standard deviations of the mean, a sample mean that corresponds to standard deviation of 2.02 from the population mean makes it seem very plausible that the people that participated in this sample weren't telling the truth. At least, the mathematics and myself, do not believe that they were telling the truth.

Hope this Helps!!!

7 0

3 years ago

An international company has 13,600 employees in one country. If this represents 19.3% of the company's employees, how many empl

user100 [1]

Hi there! To solve this question, we can write and solve a proportion. Let’s set it up like this: 13,600/x = 19.3/100. We do this, because we’re looking for the total amount of people, which would be the whole. We already have the part. Cross multiply the values in order to get 1,360,000 = 19.3x. Now, divide each side by 19.3 to isolate the variable. When you do, you get x is approx. 70,466.3212 or 70,466 when rounded to the nearest whole number. Plus, you can’t have a fraction of a person, anyway. The company has a total of 70,466 employees.

4 0

3 years ago

a square window has a perimeter of (8x + 12) feet. Write an expression that represents the side length of the window. (in feet)

Harrizon [31]

8x+12=p so you would take away 12 from p making the equation 8x=p-12 and then divide that by 8 so the equation is x= p-12/8 ( x equals p minus 12 over 8)

7 0

3 years ago