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2 years ago

7

Five teachers were hired at Oakdale Middle School at the beginning of the 2005 school year. The performance of each teacher was

rated by the principal at the conclusion of each school year. The performance ratings are charted below. Please use this information to answer the question.
Which teacher exhibited the most variable year-to-year performance over this 5 year span?

Jill

Ed

Ralph

Sam

1 answer:

Serjik [45]2 years ago

3 0

I think the answer is Jill

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With a perfectly balanced roulette wheel, in the long run, red numbers should turn up 18 times in 38. To test its wheel, one cas

topjm [15]

Answer:

a) more than 18/38

b) The null says that the percentage of reds in the box is 2842.105. The alternative says that the percentage of reds in the box is -0.933521

c) z= -0.933521 , p≤ z= 17.105%

d) p-value is 17%

Step-by-step explanation:

1. We are counting the number of reds, so we have a box with "1" s for reds and "0"s for others. The question is what proportion of 1's and 0' are in the box.

a) The null hypothesis is that the wheel is balanced, so the box contains 18 "1"s and 20 "0"s. Or you might state this in terms of the proportion of 1's (18/38) and 0's (20/38). The alternative hypothesis if that the wheel favors reds, so the proportion of 1's is more than 18/38.

b) Under the null hyp., we expect 6000*(18/38)= 2842.105 Reds. The SE is sqrt(6000)* SD(Box)= Sqrt(6000)*(1-0)*sqrt((18/38)*(20/38))= 38.67615. We observed2806, which is fewer than we expected. The question is , is it much fewer that we suspect foul play, or it is the usual sort of variation we expect due to chance? The test statistic is z=(obs-expected)/SE =(2806-2842.105)/38.67615= -0.933521.

c) The p-value is the probability of gettinga test statistic as extreme or morer extreme than this, which in this case means seeing a test statistic less than or equal to - 0.933521. The closest value gives a probability of (100-65.79)/2 = 17.105%

d) If we reject H0 and claim the roullette wheel is unfair, then there's a 17.105% chance we are wrong. This is a large probability of making a serious mistake, and it would be reasonable to not reject and conclude that there is insufficient evidence to suspect the wheel is unfair. (we would reject H10 if the p-value were less than 5%. Since it is 17%, we won't reject)

6 0

2 years ago

7 times as much as the sum of 1/3 and 4/5

Vlada [557]

$7( \frac{1}{3} + \frac{4}{5} )$

$=7( \frac{1(5)}{3(5)} + \frac{4(3)}{5(3)} )$

$=7( \frac{5}{15} + \frac{12}{15} )$

$=7( \frac{17}{15})$

$=\frac{7}{1} ( \frac{17}{15})$

$=\frac{119}{15}$

$= 7\frac{14}{15}$ ≈ 7.93

4 0

2 years ago

Read 2 more answers

Part A: During what interval(s) of the domain is the water balloon's height increasing?

Advocard [28]

Answer:

The answer is below

Step-by-step explanation:

The linear model represents the height, f(x), of a water balloon thrown off the roof of a building over time, x, measured in seconds: A linear model with ordered pairs at 0, 60 and 2, 75 and 4, 75 and 6, 40 and 8, 20 and 10, 0 and 12, 0 and 14, 0. The x axis is labeled Time in seconds, and the y axis is labeled Height in feet. Part A: During what interval(s) of the domain is the water balloon's height increasing? (2 points) Part B: During what interval(s) of the domain is the water balloon's height staying the same? (2 points) Part C: During what interval(s) of the domain is the water balloon's height decreasing the fastest? Use complete sentences to support your answer. (3 points) Part D: Use the constraints of the real-world situation to predict the height of the water balloon at 16 seconds.

Answer:

Part A: During what interval(s) of the domain is the water balloon's height increasing?

Between 0 and 2 seconds, the height of the balloon increases from 60 feet to 75 feet

Part B: During what interval(s) of the domain is the water balloon's height staying the same?

Between 2 and 4 seconds, the height remains the same at 75 feet. Also from 10 seconds the height of the balloon is at 0 feet

Part C: During what interval(s) of the domain is the water balloon's height decreasing the fastest?

Between 4 and 6 seconds, the height of the balloon decreases from 75 feet to 40 feet (i.e. -17.5 ft/s)

Between 6 and 8 seconds, the height of the balloon decreases from 40 feet to 20 feet (i.e. -10 ft/s)

Between 8 and 10 seconds, the height of the balloon decreases from 20 feet to 0 feet (i.e. -10 ft/s)

Hence it decreases fastest from 4 to 6 seconds

Part D: Use the constraints of the real-world situation to predict the height of the water balloon at 16 seconds

From 10 seconds, the balloon is at the ground, so it remains at the ground (0 feet) even at 16 seconds

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2 years ago

Segment GI is congruent to Segment JL and Segment GH is congruent to Segment KL. I have to prove Segment HI is congruent to Segm

madreJ [45]

Answer:

See explanation

Step-by-step explanation:

1 step: $\overline{GI}\cong \overline {JL}$ - given

2 step: $\overline{GI}\cong \overline{GH}+\overline{HI}$ - Segments Addition Postulate

3 step: $\overline{GH}+\overline{HI}\cong \overline {JL}$ - Substitution Property

4 step: $\overline {JL}\cong \overline {JK}+\overline {KL}$ - Segments Addition Postulate

5 step: $\overline{GH}+\overline{HI}\cong \overline {JK}+\overline {KL}$ - Substitution Property

6 step: $\overline{GH}\cong \overline {KL}$ - given

7 step: $\overline{GH}+\overline{HI}\cong \overline {JK}+\overline {GH}$ - Substitution Property of Equality

8 step: $\overline{HI}\cong \overline {JK}$ - Subtraction Property of Equality

3 0

2 years ago

Madison starts with a population of 1,000 amoebas that triples in size every hour for a number of hours, h. She writes the expre

gtnhenbr [62]

Answer: The meaning of each term of the Madison’s and Tyler’s expressions is mentioned below.

Step-by-step explanation:

Since, when Madison starts with a population of 1,000 amoebas that triples in size every hour for a number of hours, h.

That is, after 1 hour total number of amoebas = 3×1000 = $3^1\times 1000$

After 2 hour, total number of amoebas = 3×3000= $3^2\times 100$

After 3 hour, total number of amoebas = 3×9000= $3^3\times 1000$

similarly, after h hours, total number of amoebas,

f(h) = $3^h\times 1000$

where, 1000 is the initial population of amoeba 3 is the growth factor of population and f(h) is the population of amoeba after h hours.

Since, when Tyler starts with a population of 1 amoeba that increases 30% in size every hour for a number of hours.

That is, after 1 hour total number of amoebas = $(1+0.3)^1$

After 2 hour, total number of amoebas = $(1+0.3)^2$

After 3 hour, total number of amoebas = $(1+0.3)^3$

Similarly, after h hours, total number of amoebas,

f(h) = $(1+0.3)^h$

Where, 1 is the initial population of amoeba, 0.3 is the growth rate and 1.3 is the growth factor.

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2 years ago

Read 2 more answers