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3 years ago

Apply the distributive property to factor out the greatest common factor. 9+12n

Mathematics

2 answers:

ehidna [41]3 years ago

8 0

Answer:

3 is the GCF

3(3+4n)

ValentinkaMS [17]3 years ago

7 0

Answer:

3(3+4n)

Step-by-step explanation:

i do khan academy so i already know the answer trust me im not lying

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Kerri said that the product of 3.93 and 0.07 would be about 4 and would have two decimal places. Is she correct?

oee [108]

Answer:

No.

Step-by-step explanation:

This type of problem is actually a trick problem. The question indicates that the PRODUCT of 3.93 and 0.07 would be about 4. The question is referring to the answer of the two values Multiplied to each other.

To visualize the question, we have:

3.93 x 0.07

This would give us:

0.2751

Kerri would be correct if she was looking for the SUM of the two values.

This would be:

3.93 + 0.07

This would give us:

Therefore in trick problems like this ones, it is better to look at what exactly is being asked.

8 0

3 years ago

Read 2 more answers

An Impressionist painting increases in value

jok3333 [9.3K]

Answer:

A = $8406.6

Step-by-step explanation:

Given:

Average rate $r=9\%$

Initial cost of painting $a = \$1500$

Time $t = 20\ years$

We need to find the final amount of painting at the end of a 20-year.

Solution:

Using Exponential Growth rate formula as:

$A = a(1+r)^t$ ----------(1)

Where:

A = Final amount

a = Initial amount.

r = Rate as a decimal.

t = Time.

Now, we substitute all given values in equation 1.

$A = 1500(1+0.09)^{20}$

$A = 1500(1.09)^{20}$

Substitute $1.09^{20} = 5.60$ in above equation.

$A = 1500\times 5.60$

A = $8406.62

Therefore, value of the painting at the end of a 20-year A = $8406.6

5 0

3 years ago

The ratio of dogs to cats at a veterinarians office is 5 to 3 if there are 15 cats how many total animals are at the office

svet-max [94.6K]

Answer: 25

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A student who correctly answered 60 questions a test received a score of 75%. How many questions were on the test

Iteru [2.4K]

It would be 45 you take 60 x 75%

7 0

3 years ago

An article in The Engineer (Redesign for Suspect Wiring," June 1990) reported the results of an investigation into wiring errors

GarryVolchara [31]

Answer:

a) The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b) A sample of 408 is required.

c) A sample of 20465 is required.

Step-by-step explanation:

Question a:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Of 1600 randomly selected aircraft, eight were found to have wiring errors that could display incorrect information to the flight crew.

This means that $n = 1600, \pi = \frac{8}{1600} = 0.005$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 - 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0005$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 + 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0095$

The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b. Suppose we use the information in this example to provide a preliminary estimate of p. How large a sample would be required to produce an estimate of p that we are 99% confident differs from the true value by at most 0.009?

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

A sample of n is required, and n is found for M = 0.009. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.009 = 2.575\sqrt{\frac{0.005*0.995}{n}}$

$0.009\sqrt{n} = 2.575\sqrt{0.005*0.995}$

$\sqrt{n} = \frac{2.575\sqrt{0.005*0.995}}{0.009}$

$(\sqrt{n})^2 = (\frac{2.575\sqrt{0.005*0.995}}{0.009})^2$

$n = 407.3$

Rounding up:

A sample of 408 is required.

c. Suppose we did not have a preliminary estimate of p. How large a sample would be required if we wanted to be at least 99% confident that the sample proportion differs from the true proportion by at most 0.009 regardless of the true value of p?

Since we have no estimate, we use $\pi = 0.5$