Will award brainliest if answered within 10 minutes!

Please help, Geometry I have NO idea how to do this. Both Pictures! (#36, #37) WILL MARK BRAINLIEST!

aliya0001 [1]

1) The definition of a scalene triangle is that ALL sides are different.
Now, to prove a triangle is scalene, we need to show that all three sides output different values. Therefore, selections with only two sides certainly are insufficient. If we look at the last two choices, you can easily spot that the second to the last is correct answer by comparing the definition of a scalene triangle.

2) You will need to perform something similar to the previous questions, which is, find the distance between the three points, and see if they are all different => scalene; all the same => equilateral; or two of them are the same => isosceles.

The distance formula between two points (a, b) and (c,d) is given as:
$distance = \sqrt{ (a-c)^{2} + (b-d)^{2} }$

Just for you to check your work, the distances between three points are:
sqrt(34), sqrt(17), sqrt(37), which results a scalene triangle.

5 0

2 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

3 years ago

Can someone please help me I don’t know if you got to have to add 4 numbers or 2 numbers

Ghella [55]

Answer:

12 feet.

Step-by-step explanation:

You have to add 4 numbers.

The perimeter is the whole distance around the rectangle

= 2*4 + 2*2

= 8 + 4

= 12 feet.

3 0

3 years ago

Blank times 6/12 = 1/10

Goryan [66]

2/10 is the answer.

8 0

3 years ago

The diameter of Circle Q terminates on the circumference of the circle at (3,0) and (3,-7) . Write the equation of the circle in

Deffense [45]

Step-by-step explanation:

The center of a circle with 2 end points of a di diameter is the midpoint of the two endpoints.

The formula needed to find the minpoints is

(x,y) = (x2 + x1)/2, (y2 + y1)/2

x2 = 3

x1 = 3

y2 = 0

y1 = -7

midpoint = (3 + 3)/2, (0 - 7)/2

midp[oint = 3,-3.5

The midpoint is the center of the circle. Observe that the signs get changed when entering the values for (x,y)

So far what you have is (x - 3)^2 + (y + 3.5)^2 = r^2

To determine r^2 you need only take the distance from the center to oneof the endpoints.

r^2 = (3 - 3)^2 + (3.5 - 0)^2

r^2 = 3.5^2

r^2 = 12.25

Answer: (x - 3)^2 + (y + 3.5)^2 = 12.25

6 0

2 years ago