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3 years ago

Graph the set of points. Which model is most appropriate for the set? (-6 -1) (-3,2) (-1 4) (2 7)

Mathematics

1 answer:

Strike441 [17]3 years ago

5 0

These are the points. To see if there is a change in their slope, we need to look at each point.

The slope between (-6,-1) and (-3,2) is
$\frac{ - 6 - ( - 3)}{ - 1 - 2} \\ \frac{ - 3}{ - 3} \\ = 1$
The slope between (-3,2) and (-1,4) is
$\frac{ - 3 - ( - 1)}{2 - 4} \\ \frac{ - 2}{ - 2} \\ = 1$
The slope between (-1,4) and (2,7) is
$\frac{ - 1 - 2}{4 - 7} \\ \frac{ - 3}{ - 3} \\ = 1$

Since the slopes are exactly the same, these points form a line. The answer is A. linear.

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Please help! Solve x/12+1/3=−5/3. What is x?

Mama L [17]

the answer would be: x= -24

6 0

3 years ago

Read 2 more answers

Determine whether each scale factor produces an expansion or a contraction under a dilation of the original image

blondinia [14]

Answer:

see the explanation

Step-by-step explanation:

we know that

If the absolute value of the scale factor is less than 1, then the dilation produces a contraction of the original image

If the absolute value of the scale factor is greater than 1, then the dilation produces an expansion of the original image

<u><em>Verify each value</em></u>

1) -4

$\left|-4\right|=4$

$4> 1$

therefore

The dilation produces an expansion of the original image

2) 0.25

$\left|0.25\right|=0.25$

$0.25 < 1$

therefore

The dilation produces a contraction of the original image

3) -2/3

$\left|-2/3\right|=2/3$

$2/3 < 1$

therefore

The dilation produces a contraction of the original image

4) 2.3

$\left|2.3\right|=2.3$

$2.3 > 1$

therefore

The dilation produces an expansion of the original image

4 0

3 years ago

Triangle XYZ was dilated by a scale factor of 2 to create triangle ACB and sin ∠X = 5 over 5 and 59 hundredths.

Vikentia [17]

The relationship between Traingle XYZ and ACB is that they are similar triangles. tanX = tanA = 5 over 2 and 5 tenths, where

AC = 2 x XY
CB = 2 x YZ

<h3>What is the dilation about?</h3>

Triangle XYZ was dilated by a scale factor of 2 to create triangle ACB, hence XYZ and ACB are similar triangles.

Angles Y and C are said to measure 90 degrees, and angles A and X are known to be congruent. Thus:

tanX = tanA = 5 over 2 and 5 tenths

AC = 2 x XY

CB = 2 x YZ

Another way to solve for it is by:

Note that Dilation of the triangle ΔXYZ was by a factor of "2".

Since m∠Y = m∠C = 90º - given

ΔXYZ and ΔACB are said to be right triangles.

Since ∠X ≅ ∠A - given

Then ∠X and ∠Z are said to becomplementary angles

Since m∠Z = 90° - m∠X -- given

Then ∠A and ∠B are said to be complementary angles

Since m∠B = 90° - m∠A --- given

Then, ∠Z ≅ ∠B

Therefore ΔXYZ ∼ ΔACB are similar triangles because it has its corresponding sides to be proportional and its corresponding angles to be congruent.

Note:

sin ∠X = 5/5.59

sin ∠X = YZ/XY

YZ = 5 seen in ΔXYZ

XZ = 5.59 seen in hypotenuse of ΔXYZ

Then one need to Calculate the length of one aspect of XY:

XY = √((5.59)2 - 52)

= 2.4996

Note CB/YZ = 2

CD = 2*YZ = 2 x 5

= 10

AC/XY = 2

AC = 2* XY

= 2 x 2.4996

= 4.999

Learn more about dilation from:

brainly.com/question/27517432

#SPJ1

7 0

2 years ago

Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviati

Gennadij [26K]

Answer:

a. the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, the probability that they have pulse rates with a mean less than 76 beats per minute is 0.8849

c. D. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Given that:

Mean μ =73.0

Standard deviation σ =12.5

a. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 76 beats per minute.

Let X represent the random variable that is normally distributed with a mean of 73.0 beats per minute and a standard deviation of 12.5 beats per minute.

Then : X $\sim$ N ( μ = 73.0 , σ = 12.5)

The probability that her pulse rate is less than 76 beats per minute can be computed as:

$P(X < 76) = P(\dfrac{X-\mu}{\sigma}< \dfrac{X-\mu}{\sigma})$

$P(X < 76) = P(\dfrac{76-\mu}{\sigma}< \dfrac{76-73}{12.5})$

$P(X < 76) = P(Z< \dfrac{3}{12.5})$

$P(X < 76) = P(Z< 0.24)$

From the standard normal distribution tables,

$P(X < 76) = 0.5948$

Therefore , the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 76 beats per minute.

now; we have a sample size n = 25

The probability can now be calculated as follows:

$P(\overline X < 76) = P(\dfrac{\overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{ \overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}})$

$P( \overline X < 76) = P(\dfrac{76-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{76-73}{\dfrac{12.5}{\sqrt{25}}})$

$P( \overline X < 76) = P(Z< \dfrac{3}{\dfrac{12.5}{5}})$

$P( \overline X < 76) = P(Z< 1.2)$

From the standard normal distribution tables,

$P(\overline X < 76) = 0.8849$

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

In order to determine the probability in part (b); the normal distribution is perfect to be used here even when the sample size does not exceed 30.

Therefore option D is correct.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

5 0

3 years ago

When we use the SPDC model, we always include the following EXCEPT:

Sidana [21]

B.!!!!!! !!!!
that’s the answer

4 0

3 years ago