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4 years ago

6

Bart and Sam played a game in which each player earns or loses points in each turn. A player's total score after two turns is th

e sum of his points earned or lost. The player with the greater score after two turns wins. Bart earned 123 points and lost 180 points. Sam earned 185 points and lost 255 points. Which person won the game? Explain.

2 answers:

lina2011 [118]4 years ago

7 0

Answer:

Step-by-step explanation:

Bart 123-180=-57

Sam 185-255= -70

Bart wins bc -57 is greater than -70

Hatshy [7]4 years ago

6 0

Bart 123-180=-57
Sam 185-255= -70
Bart wins bc -57 is greater than -70

You might be interested in

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.9

Eva8 [605]

Answer:

a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes

c) 0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 8.9 minutes and a standard deviation of 2.9 minutes.

This means that $\mu = 8.9, \sigma = 2.9$

Sample of 37:

This means that $n = 37, s = \frac{2.9}{\sqrt{37}}$

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

320/37 = 8.64865

Sample mean below 8.64865, which is the p-value of Z when X = 8.64865. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{8.64865 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -0.53$

$Z = -0.53$ has a p-value of 0.2981

0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

275/37 = 7.4324

Sample mean above 7.4324, which is 1 subtracted by the p-value of Z when X = 7.4324. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{7.4324 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -3.08$

$Z = -3.08$ has a p-value of 0.001

1 - 0.001 = 0.999

0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Sample mean between 7.4324 minutes and 8.64865 minutes, which is the p-value of Z when X = 8.64865 subtracted by the p-value of Z when X = 7.4324. So

0.2981 - 0.0010 = 0.2971

0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

7 0

3 years ago

3. What three numbers are the Pythagorean triple generated by using 4 for x and 1 for y. Remember to use the formulas:

Gnom [1K]

Given:

The three equations for the three numbers are

$a=x^2-y^2$

$b=2xy$

$c=x^2+y^2$

To find:

The Pythagorean triple generated by using 4 for x and 1 for y.

Solution:

Substituting x=4 and and y=1 in the given equations, we get

$a=4^2-1^2$

$a=16-1$

$a=15$

In the same way find b and c.

$b=2(4)(1)$

$b=8$

The third number is

$c=4^2+1^2$

$c=16+1$

$c=17$

The required Pythagorean triple is 8, 15, and 17.

Therefore, the correct option is C.

5 0

3 years ago

Can someone please help me!!

Elan Coil [88]

Answer:

13

Step-by-step explanation:

Firstly we would start off by simplifying the equation by doing the order of operations backward (PEMDAS -> SADMEP) and multiply -4 and -3 which would equal a positive 12

Next, we would divide 12 and -6 and get a negative 2

Finally, we would have the equation 1 - 12 + (-2)

We would solve the equation to get -13/(-1)³

This would make the -13 into a positive since any number over one is a whole number. -13 is getting divided by (-1) 3 times so it converts to a positive, then a negative, then back to a positive.

I hope this helped :3

6 0

3 years ago

A car moves at a constant speed of 90km/h from a starting point. Another car moves at 70km/h after 2hours from the same starting

patriot [66]

Step-by-step explanation:

are you sure you wrote the problem here correctly ?

because the distance will be 40km after less than half an hour just by the first car driving. way before the second car even starts.

to be precise, it would be after 60 minutes × 40 / 90

(= how many minutes of an hour are needed to reach 40km while going 90km/h) :

60 × 40 / 90 = 60 × 4 / 9 = 20 × 4 / 3 = 80/3 = 26.67 minutes.

but maybe the question was about 400km distance between the two cars.

so, the first car goes 90km/h for 2 hours.

at that moment it will be 2×90=180km ahead.

that would mean that 220km are still missing for the 400km assumption.

with each hour driving the first car makes 20km more than the second car.

to build up 220km that way would require

220/20 = 11 hours.

plus the 2 original head start hours this would make 13 hours as overall answer.

7 0

3 years ago

Pls help being timed

Vera_Pavlovna [14]

4x = 300*99

x= 300*99/4 = 7425.

So a is correct asnwer

6 0

3 years ago

Read 2 more answers