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3 years ago

Find the distance between the following points C(11, -12), D(6, 2)

Mathematics

1 answer:

Lady bird [3.3K]3 years ago

3 0

Answer:

14.9

Step-by-step explanation:

Formula= $\sqrt{(x2-x1)^{2} +(y2-y1)^{2}$

x1=11; y1=-12; x2=6; y2=2

Putting these values in the formula, the result will be 14.866≡14.9

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Will mark brainliest

VMariaS [17]

Answer:

Interquartile range => 7

Third quartile => 22

Range => 13

First quartile => 15

Step-by-step explanation:

Order your data set, from the least amount to the highest amount:

$12, $14, $15, $15, $15, $15, $16, $22, $24, $25

Interquartile range (IQR) = third quartile (Q3) - first quartile (Q1)

Q1 = the middle value of the lower part of the data set, from the median to your left.

Q3 = the middle value of the upper part of the data set, from the median to your right.

The median lies between the 5th and 6th value that is enclosed in the parenthesis below:

$12, $14, ($15), $15, $15,[Median], $15, $16, ($22), $24, $25

The median divides the data set into upper and lower part.

Median = $\frac{15 + 15}{2} = 15$

First quartile: Q1 = $15

Third quartile: Q3 = $22

IQR = $22 - $15 = $7

Range = highest amount - least amount = 25 - 12 = $13

3 0

3 years ago

Help plz:))) I’ll mark u brainliest

egoroff_w [7]

Answer:

Im pretty sure it is c

Step-by-step explanation:

it gives us 2 angles add up to 180, but they are right next to each other. So we cant determine parallel lines.

Can i have brainliest now? im trying to work up the ranks

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3 years ago

The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probabili

Lena [83]

Answer:

a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

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 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.

The mean of a population is 74 and the standard deviation is 15.

This means that $\mu = 74, \sigma = 15$

Question a:

Sample of 36 means that $n = 36, s = \frac{15}{\sqrt{36}} = 2.5$

This probability is 1 subtracted by the pvalue of Z when X = 78. So

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

By the Central Limit Theorem

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

$Z = \frac{78 - 74}{2.5}$

$Z = 1.6$

$Z = 1.6$ has a pvalue of 0.9452

1 - 0.9452 = 0.0548

0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

Question b:

Sample of 150 means that $n = 150, s = \frac{15}{\sqrt{150}} = 1.2247$

This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So

X = 77

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

$Z = \frac{77 - 74}{1.2274}$

$Z = 2.45$

$Z = 2.45$ has a pvalue of 0.9929

X = 71

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

$Z = \frac{71 - 74}{1.2274}$

$Z = -2.45$

$Z = -2.45$ has a pvalue of 0.0071

0.9929 - 0.0071 = 0.9858

0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c. A random sample of size 219 yielding a sample mean of less than 74.2

Sample size of 219 means that $n = 219, s = \frac{15}{\sqrt{219}} = 1.0136$

This probability is the pvalue of Z when X = 74.2. So

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

$Z = \frac{74.2 - 74}{1.0136}$

$Z = 0.2$

$Z = 0.2$ has a pvalue of 0.5793

0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

5 0

3 years ago

Betty's grade in science class is affected by the number of missing assignments. The table below shows the science grade percent

Angelina_Jolie [31]

Hi!

$y=-5x+95$

<em>Plug in </em> $15$ for $x$

$y=-5(15)+95$

<em>Multiply </em> $-5$ <em>and </em> $15$

$y=-75+95$

<em>Add </em> $-75$ <em>and </em> $95$

$y =15$

-------------------------------------------------------------------------------------------------------------

Betty's percentage is 15%.

The rate of change is the amount it changes for every missing assignment, therefore it would be -5.

5 0

3 years ago

(1) Tom took a trip of 1,020 miles. He traveled by train at 55 miles an hour and the same number of hours by plane at 285 mph.

Maurinko [17]

<h3>Answer:</h3>

B) 6 hours

<h3>Step-by-step explanation:</h3>

Let h represent the total trip time. Then h/2 is the time spent traveling at each speed. The distance covered is ...

... distance = speed · time

The sum of the distances in each mode is the total distance.

... 1020 = 55·h/2 +285·h/2

... 2040 = h·(55+285) . . . . multiply by 2

... 2040/340 = h = 6 . . . . . . . divide by the coefficient of h

The trip took a total of 6 hours.

4 0

3 years ago