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2 years ago

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A survey asked eight people about their wages and educational background. The table shows the hourly wages reported by people wi

th and without a high school diploma. A 2-column table with 4 rows. Column 1 is labeled No high school diploma with entries 10.00, 9.50, 11.50, 13.00. Column 2 is labeled High School Diploma with entries 19.00, 15.25, 14.00, 15.75. Use the information to complete the statements. The mean absolute deviation for people without a high school diploma is . The mean absolute deviation for people with a high school diploma is . The data for people without a high school diploma are more the mean than the data for people with a high school diploma.

2 answers:

svet-max [94.6K]2 years ago

7 0

Answer:The first one is 1.25 the second one is 1.5 ald the 3 one is clustered around

Step-by-step explanation:

it is very easy

olganol [36]2 years ago

5 0

Answer:

1:1.25 2: 1.5 3: clusterd around

Step-by-step explanation:

i am smarte

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The percent defective for parts produced by a manufacturing process is targeted at 4%. The process is monitored daily by taking

Anna [14]

Answer:

The 88% confidence interval for the proportion of defectives today is (0.053, 0.123)

Step-by-step explanation:

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 $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 160, \pi = \frac{14}{160} = 0.088$

88% confidence level

So $\alpha = 0.12$ , z is the value of Z that has a pvalue of $1 - \frac{0.12}{2} = 0.94$ , so $Z = 1.555$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.088 - 1.555\sqrt{\frac{0.088*0.912}{160}} = 0.053$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.088 + 1.555\sqrt{\frac{0.088*0.912}{160}} = 0.123$

The 88% confidence interval for the proportion of defectives today is (0.053, 0.123)

6 0

1 year ago

A library buys 36 English books, 48 Science books and 72 Mathematics books. The thickness of each book is the same. Now, the lib

fomenos

Answer:

13

Step-by-step explanation:

The GCF of 36, 48, and 72 is 12 so there will be 36 / 12 = 3 stacks of English books, 48 / 12 = 4 stacks of science books and 72 / 12 = 6 stacks of math books for a total of 3 + 4 + 6 = 13 stacks.

5 0

2 years ago

Read 2 more answers

Eight times the reciprocal of a number equals 2 times the reciprocal of 9. Find the number

vfiekz [6]

Answer:

8 * 1/x = 1/9 *2

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1 year ago

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

Jeanie completed a 27 mile duathlon (a race that is a combination of running and biking) in exactly 2 hours. she ran an average

-Dominant- [34]

<span>Let x = the miles Jeanie ran y = the miles Jeanie biked T = 2 hours total time for the duathlon v1 = 8.5 mph Jeanie's running speed and v2 = 16 mph Jeanie's biking speed But T = Jean's time she ran t1 + time she biked t2 I. e t1 + t2 = 2 So we have average speed v2 when she biked = distance (y) / time (t2) Distance y = v2 * t2 = 16 * t2 and distance when she ran x = 8.5 * t1 Since she covered 27 miles while running and biking we have x + y = 27 8.5t1 + 16t2 =27 ------(1) t1 + t2 = 2 --------------(2) The simultaneous equation gives us t1 = 2/3 and t2 which is time she biked = 4/3 So 4/3 = 1 1/3. Which is 1 hour 20 minutes</span>

8 0

2 years ago