Write the polynomial as a product: mn − mk + xk − xn

Ming writes a number pattern on a slip of paper and hands it to his friend Jack. 24 21 23 20 22 19 21 18 Jack writes the next nu

tiny-mole [99]

The rule is -3, then +2

24-3 = 21
21 + 2 = 23
23-3= 20
20+2= 22
22-3= 19
19+2 = 21
21-3= 18
18+2 = 20

3 0

1 year ago

HEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEELLLLLLLLLLLLLLLLLLLLLLLLP What is the simplified value of the expression below?

olganol [36]

Answer:

100

Step-by-step explanation:

start with what's in the parenthesis':

8 (9.75 - 3.25) + 12 x 4 = 8 (6.50) + 12 x 4

then take the number outside of the parenthesis' time the number inside of them:

8 (6.50) + 12 x 4 = 52 + 12 x 4

now, take the product of the two number on the right side of the addition sign:

52 + 12 x 4 = 52 + 48

finally, add the final two numbers together:

52 + 48 = 100

6 0

1 year ago

Read 2 more answers

What proportion results in the equation 9m=10n

Serhud [2]

Give the equation 9m = 10n

Dividing both sides by 9n, we have
$\frac{9m}{9n} = \frac{10n}{9n} \\ \\ \frac{m}{n} = \frac{10}{9}$

Therefore, the <span>proportion that results in the equation 9m = 10n is
$\frac{m}{n}=\frac{10}{9}$
</span>

4 0

2 years ago

Read 2 more answers

The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat

Greeley [361]

Answer:

1) $L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

2) $T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

3) $T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

$L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

Part 2

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

Part 3

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

Replacing we got:

$T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

8 0

2 years ago

Customers arrive at a movie theater at the advertised movie time only to find that they have to sit through several previews and

lubasha [3.4K]

This question is incomplete, the complete question is;

Customers arrive at a movie theater at the advertised movie time only to find that they have to sit through several previews and preview ads before the movie starts. Many complain that the time devoted to previews is too long (The Wall Street Journal, October 12, 2012). A preliminary sample conducted by The Wall Street Journal showed that the standard deviation of the amount of time devoted to previews was 4 minutes. Use that as a planning value for the standard deviation in answering the following questions. Round your answer to next whole number,

a) If we want to estimate the population mean time for previews at movie theaters with a margin of error of 75 seconds, what sample size should be used? Assume 95% confidence.

b) If we want to estimate the population mean time for previews at movie theaters with a margin of error of 1 minute, what sample size should be used? Assume 95% confidence.

Answer:

a) the required Sample Size is 40

b) the required Sample Size is 62

Step-by-step explanation:

Given the data in the question;

standard deviation σ = 4 minutes

a)

margin of error E = 75 seconds = ( 75 / 60 )minutes = 1.25 minutes

And 95% confidence.

Now, Critical Value of z for 0.95 confidence interval;

∝ = 1 - 0.95 = 0.05

$Z_{\alpha /2$ = 1.96

so, sample size n will be;

n = [ $Z_{\alpha /2$ × σ/E ]²

we substitute;

n = [ 1.96 × (4/1.25) ]²

n = [ 1.96 × 3.2 ]²

n = [ 6.272 ]²

n = 39.338

Since we referring to a number of sample, its approximately becomes 40

Therefore, the required Sample Size is 40

b)

margin of error E = 1 minutes

And 95% confidence.

Now, Critical Value of z for 0.95 confidence interval;

∝ = 1 - 0.95 = 0.05

$Z_{\alpha /2$ = 1.96

so, sample size n will be;

n = [ $Z_{\alpha /2$ × σ/E ]²

we substitute;

n = [ 1.96 × (4/1) ]²

n = [ 1.96 × 4 ]²

n = [ 7.84 ]²

n = 61.466

Since we referring to a number of sample, its approximately becomes 62

Therefore, the required Sample Size is 62

6 0

1 year ago