All categories

3 years ago

What is the slope of the line that contains the points (-2,5) and (6-3)

Mathematics

1 answer:

sashaice [31]3 years ago

3 0

Answer: m = -1

Step-by-step explanation: In this problem we're asked to find the slope of the line that contains the points (-2,5) and (6,-3).

Using our slope formula, we take the second y minus the first y which in this case is -3 - 5 over our second x minus our first x which in this case is 6 - -2.

-3 - 5 simplifies to -8 and remember that minus a negative can be thought of as plus a positive so 6 - (-2) can be thought of as 6 + (+2) which is 8.

Now we have -8/8. Notice that our slope can be simplified one step further so we have m = -1.

The variable that's used to represent slope is m.

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120 deg

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Step-by-step explanation:

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2 years ago

-52 times the difference between a number and 14 is equal to the number plus 3

xenn [34]

X=number looked for
we have this equation:
-52(x-14)=x+3
we solve this equation:
-52x+728=x+3
-52x-x=3-728
-53x=-725
x=-725/-53=725/53 ≈13,68

Solution: 725/53

To check:
the difference between 725/53 and 14=725/53 - 14=(725-742)/53=-17/53
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3 years ago

A student records the number of hours that they have studied each of the last 23 days. They compute a sample mean of 2.3 hours a

natita [175]

Answer:

the standard deviation increases

Step-by-step explanation:

Let x₁ , x₂, . . . , x₂₃ be the actual data observed by the student

The sample means = x₁ + x₂ + . . . , x₂₃ / 23

$= \frac{x_1 +x_2 +...x_2_3}{23}$

= 2.3hr

⇒ $\sum xi =2.3 \times 23 = 52.9hrs$

let x₁ , x₂, . . . , x₂₃ arranged in ascending order

Then x₂₃ was 10 and has been changed to 14

i.e x₂₃ increase to 4

Sample mean $= \frac{x_1 +x_2 +...x_2_3}{23}$

$\frac{52.9hrs + 4}{23} \\\\= \frac{56.9}{23} \\\\= 2.47$

therefore, the new sample mean is 2.47

2) For the old data set

the median is $x_1_2(th)$ values

$[\frac{n +1}{2} ]^t^h value$

when we use the new data set only x₂₃ is changed to 14

i.e the rest all observation remain unchanged

Hence, sample median = $[{x_1_2]^t^h value$ remain unchange

sample median = 2.5hrs

The Standard deviation of old data set is calculated

$=\sqrt{\frac{1}{n-1} \sum (xi - \bar x_{old})^2 } \\\\=\sqrt{\frac{1}{22}\sum ( xi - 2.3)^2 }---(1)$

The new sample standard sample deviation is calculated as

$= \sqrt{\frac{1}{n-1} \sum (xi-2.47)^2} ---(2)$

Now, when we compare (1) and (2) the square distance between each observation xi and old mean is less than the squared distance between each observation xi and the new mean.

Since,

(xi - 2.3)² ∑ (xi - 2.47)²

Therefore , the standard deviation increases

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