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adelina 88 [10]
3 years ago
13

X+2(x+y)=7 3(x-y)=1-2y What is x&y???? Please answer asap!!!!

Mathematics
1 answer:
Shkiper50 [21]3 years ago
4 0
X= -x over 2 (negative x as a fraction and 2 is the denominator) minus y+ (seven as the numerator over 2 as the denominator).
Solve for y by simplify both sides with inverse operations, then isolate the variable and you get,  y = 3x - 1.
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The time a randomly selected individual waits for an elevator in an office building has a uniform distribution with a mean of 0.
Amiraneli [1.4K]

Answer:

The mean of the sampling distribution of means for SRS of size 50 is \mu = 0.5 and the standard deviation is s = 0.0409

By the Central Limit Theorem, since we have of sample of 50, which is larger than 30, it does not matter that the underlying population distribution is not normal.

0% probability a sample of 50 people will wait longer than 45 seconds for an elevator.

Step-by-step explanation:

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

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size, of at least 30, can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 0.5, \sigma = 0.289

What are the mean and standard deviation of the sampling distribution of means for SRS of size 50?

By the Central Limit Theorem

\mu = 0.5, s = \frac{0.289}{\sqrt{50}} = 0.0409

The mean of the sampling distribution of means for SRS of size 50 is \mu = 0.5 and the standard deviation is s = 0.0409

Does it matter that the underlying population distribution is not normal?

By the Central Limit Theorem, since we have of sample of 50, which is larger than 30, it does not matter that the underlying population distribution is not normal.

What is the probability a sample of 50 people will wait longer than 45 seconds for an elevator?

We have to use 45 seconds as minutes, since the mean and the standard deviation are in minutes.

Each minute has 60 seconds.

So 45 seconds is 45/60 = 0.75 min.

This probability is 1 subtracted by the pvalue of Z when X = 0.75. So

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

By the Central Limit Theorem

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

Z = \frac{0.75 - 0.5}{0.0409}

Z = 6.11

Z = 6.11 has a pvalue of 1

1-1 = 0

0% probability a sample of 50 people will wait longer than 45 seconds for an elevator.

8 0
3 years ago
A paper cup is tossed 50 times. It lands right side up 11 times, upside down 14 times, and on its side 25 times. What percent of
Lynna [10]

Answer

28 percent

Step-by-step explanation:

divide 14/50

3 0
3 years ago
Sb pls help me with this question
11111nata11111 [884]
B is true; pi= ~3.14. 3+4 is equal to 7, so because pi is greater than 3, pi+4>7
8 0
2 years ago
A ball is thrown into the air with a velocity of 34 ft/s. It's height in feet after t seconds is given by y=34t-26(t)^2. Find th
kkurt [141]

Answer:

The average velocity of the ball at the given time interval is -122.3 ft/s

Step-by-step explanation:

Given;

velocity of the ball, v = 34 ft/s

height of the ball, y = 34t - 26t²

initial time, t₀ = 3 seconds

final time, t = 3 + 0.01 = 3.01 seconds

At t = 3 s

y(3) = 34(3) - 26(3)² = -132

The average velocity of the ball in ft/s is given as;

V_{avg} = \frac{y(3.01)-y(3)}{3.01 -3}\\\\V_{avg} = \frac{34(3.01)-26(3.01)^2-y(3)}{3.01 -3}\\\\V_{avg} = \frac{-133.223-y(3)}{0.01}\\\\V_{avg} = \frac{-133.223-(-132)}{0.01}\\\\V_{avg} =\frac{-1.223}{0.01}\\\\V_{avg} = -122.3 \ ft/s

Therefore, the average velocity of the ball at the given time interval is -122.3 ft/s

8 0
3 years ago
Which fraction names the shaded part of the set?
Mademuasel [1]
It would be 4/12 which is equal to 1/3.

The answer is 1/3
7 0
2 years ago
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