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Dennis_Churaev [7]
3 years ago
12

Which of the following ordered pairs represents the solution to the system given below?

Mathematics
1 answer:
pentagon [3]3 years ago
4 0
(3,-1) just fill it in and the only one that makes sense is the first one 
You might be interested in
How many meters tall is the statue of liberty if it is thirty times the height of a 154 centimeter person?
Reptile [31]
The question states that the Statue of Liberty is 30 times the height of a 154 centimeter person and asks how many meters tall the <span>the Statue of Liberty is.
This is basically asking us to find 30 times 154 centimeters and convert it to meters.

30 • 154 = 4620
This tells us that the </span>Statue of Liberty is 4,620 centimeters (cm) tall.

Now we must convert 4,620 cm to meters (m).
There are 100 cm in 1 m.
This means 100 cm = 1 m.
That means that meters are 100 times larger than centimeters.
With this in mind, we can divide the number of cm by 100 to convert it to m.

4,620 ÷ 100 = 46.2
That means that 4,620 cm is equal to 46.2 m.

The final answer:
If the Statue of Liberty is 30 times taller than 154 centimeters, then the Statue of Liberty is 46.2 meters tall.
So the answer is 46.2 meters.

Hope this helps!
8 0
3 years ago
Read 2 more answers
Really quick question please answer it in fraction form. What is 6* 2/5?
sdas [7]

Answer:

12/5

Step-by-step explanation:

6(2/5)

= (6/1) * (2/5)

= (6)(2) / (1)(5)

= 12/5

7 0
3 years ago
The product of two consecutive positive even integers is 440. what are the integers
Natali [406]
1st integer: x
2nd integer: x+2 

Multiply them: x(x+2)=440
X^2+2x-440=0 
Use the quadratic formula 
x=-22 or x=20 
Since x must be a positive integer number, x=20.

Answer: 
1st: 20.
2nd: 22 

Rate High and send a personal message if you want detailed answers on your math questions.
4 0
3 years ago
The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation
fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

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 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 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.

In this question, we have that:

\mu = 197.5, \sigma = 8.3

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

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

Z = \frac{210.9 - 197.5}{8.3}

Z = 1.61

Z = 1.61 has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

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

By the Central Limit Theorem

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

Z = \frac{196 - 197.5}{2.14}

Z = -0.7

Z = -0.7 has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

5 0
3 years ago
Write 24.652 as a mixed number.
Vladimir79 [104]
24.652 = 24   65/100


24  65/100 is your answer

hope this helps

4 0
3 years ago
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