Help I don’t get this please I need help

What is the correct answer?

emmasim [6.3K]

The answer is option 2

8 0

3 years ago

Read 2 more answers

AAST is dilated about the origin o to create ARST. Point Ris located at (3.9) and point R’ is

Slav-nsk [51]

The scale factor which used to perform the dilation was 0.4

Step-by-step explanation:

If point (x , y) is dilated about the origin by a scale factor k, then

Its image is point (kx , ky)
If k > 1, then kx > x and ky > y
If 0 < k < 1, then kx < x and ky < y

∵ Δ RST is dilated about the origin O to create Δ R'S'T'

- Assume that the scale factor is k

∵ Point R is (3 , 9)

∴ x = 3 and y = 9

∵ Point R' is (1.2 , 3.6)

∴ kx = 1.2 and ky = 3.6

∵ x = 3 and kx = 1.2

- Substitute x in kx to find k

∴ k(3) = 1.2

∴ 3k = 1.2

- Divide both sides by 3

∴ k = 0.4

To check your answer use the y-coordinates of R and R' to find k

∵ y = 9 and ky = 3.6

- Substitute the value of y in ky to find k

∵ k(9) = 3.6

∴ 9k = 3.6

- Divide both sides by 9

∴ k = 0.4

∴ The scale factor is 0.4

The scale factor which used to perform the dilation was 0.4

Learn more:

You can learn more about dilation in brainly.com/question/3552621

#LearnwithBrainly

5 0

3 years ago

AB is tangent to the circle at B. <A=14. Arc BC = 112.

goldenfox [79]

So hmm check the picture below

a circle has a total of 360°, thus arcBC + x + arcCD is 360, so arcCD is just 360 - 112 - x, whatever "x" was that you found

3 0

3 years ago

Find all points on the y-axis that are 5 units from the point<br> (4,4).

V125BC [204]

Answer:

-4,9 , -4,-1 , 4,-1 , 4,9

Step-by-step explanation:

The y-axis is straight up and if you move down on the and moving up 5 is 9 and moving down 5 is -1 and depending if it's on the right of left (negative or positive) the x-axis is (the 4) is postive or negative.

3 0

3 years ago

The following table shows scores obtained in an examination by B.Ed JHS Specialism students. Use the information to answer the q

Makovka662 [10]

Answer:

(a) The cumulative frequency curve for the data is attached below.

(b) (i) The inter-quartile range is 10.08.

(b) (ii) The 70th percentile class scores is 0.

(b) (iii) the probability that a student scored at most 50 on the examination is 0.89.

Step-by-step explanation:

(a)

To make a cumulative frequency curve for the data first convert the class interval into continuous.

The cumulative frequencies are computed by summing the previous frequencies.

The cumulative frequency curve for the data is attached below.

(b)

(i)

The inter-quartile range is the difference between the third and the first quartile.

Compute the values of Q₁ and Q₃ as follows:

Q₁ is at the position:

$\frac{\sum f}{4}=\frac{100}{4}=25$

The class interval is: 34.5 - 39.5.

The formula of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 25 = 34.5

(CF) $_{p}$ = cumulative frequency of the previous class = 24

f = frequency of the class interval = 20

h = width = 39.5 - 34.5 = 5

Then the value of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

$=34.5+[\frac{25-24}{20}]\times5\\\\=34.5+0.25\\=34.75$

The value of first quartile is 34.75.

Q₃ is at the position:

$\frac{3\sum f}{4}=\frac{3\times100}{4}=75$

The class interval is: 44.5 - 49.5.

The formula of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 75 = 44.5

(CF) $_{p}$ = cumulative frequency of the previous class = 74

f = frequency of the class interval = 15

h = width = 49.5 - 44.5 = 5

Then the value of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

$=44.5+[\frac{75-74}{15}]\times5\\\\=44.5+0.33\\=44.83$

The value of third quartile is 44.83.

Then the inter-quartile range is:

$IQR = Q_{3}-Q_{1}$

$=44.83-34.75\\=10.08$

Thus, the inter-quartile range is 10.08.

(ii)

The maximum upper limit of the class intervals is 69.5.

That is the maximum percentile class score is 69.5th percentile.

So, the 70th percentile class scores is 0.

(iii)

Compute the probability that a student scored at most 50 on the examination as follows:

$P(\text{Score At most 50})=\frac{\text{Favorable number of cases}}{\text{Total number of cases}}$

$=\frac{10+4+10+20+30+15}{100}\\\\=\frac{89}{100}\\\\=0.89$

Thus, the probability that a student scored at most 50 on the examination is 0.89.

5 0

4 years ago