All categories

3 years ago

In Mr. Diaz’s class, 20% of the students received a grade of “A” on their report cards. There are 25 students in

Mathematics

2 answers:

Simora [160]3 years ago

7 0

25/x=100/20 criss cross and u will get 500/100=5 then 5 students resivs A.

alexandr1967 [171]3 years ago

7 0

Im not sure but I think 5 because there are 25 students so put 25 in the denominator Set up a proportion Put 20 over 100 because percent means something over 100 25 times 4 is 100 20 divided by 4 is 5 So the answer is 5

You might be interested in

The actual length of side t is 0.045 cm. Use the scale drawing to find the actual side length of w. A) 0.06 cm B) 0.075 cm C) 0.

Tems11 [23]

Answer: The answer is B 0.075 cm

Step-by-step explanation:

Since the drawings is a scale of the original you must use a proportion to solve the desired side. 0.045/.9=w/1.5;w=0.075.

5 0

3 years ago

Read 2 more answers

(show work)

bearhunter [10]

Answer:

JM is the diameter or 180°

MK = diameter - 21° = 159°

Step-by-step explanation:

8 0

2 years ago

NEED HELP ASAP PLEASE WITH #7!!

AnnZ [28]

Answer:

The answer to your question is: The second option

Step-by-step explanation:

Points A (3 , 7) B (5, 11)

slope = m = (y2 - y1) / (x2 - x1)

m = (11 - 7) / (5 - 3)

m = 4 / 2 = 2

equation if the line

(y - y1) = m(x - x1)

(y - 7) = 2(x - 3) Answer

6 0

3 years ago

Find the measures of the angles of the triangle whose vertices are A = (-3,0) , B = (1,3) , and C = (1,-3).A.) The measure of ∠A

alekssr [168]

Answer:

$\theta_{CAB}=128.316$

$\theta_{ABC}=25.842$

$\theta_{BCA}=25.842$

Step-by-step explanation:

A = (-3,0) , B = (1,3) , and C = (1,-3)

We're going to use the distance formula to find the length of the sides:

$r= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

$AB= \sqrt{(-3-1)^2+(0-3)^2}=5$

$BC= \sqrt{(1-1)^2+(3-(-3))^2}=9$

$CA= \sqrt{(1-(-3))^2+(-3-0)^2}=5$

we can use the cosine law to find the angle:

it is to be noted that:

the angle CAB is opposite to the BC.

the angle ABC is opposite to the AC.

the angle BCA is opposite to the AB.

to find the CAB, we'll use:

$BC^2 = AB^2+CA^2-(AB)(CA)\cos{\theta_{CAB}}$

$\dfrac{BC^2-(AB^2+CA^2)}{-2(AB)(CA)} =\cos{\theta_{CAB}}$

$\cos{\theta_{CAB}}=\dfrac{9^2-(5^2+5^2)}{-2(5)(5)}$

$\theta_{CAB}=\arccos{-\dfrac{0.62}}$

$\theta_{CAB}=128.316$

Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

$\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}$

$\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)$

$\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)$

$\theta_{ABC}=\arcsin{0.4359}\right)$

$\theta_{ABC}=25.842$

finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

$\theta_{BCA}=25.842$

this can also be checked using the fact the sum of all angles inside a triangle is 180

$\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180$