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telo118 [61]
3 years ago
12

Last Sunday 1,575 people visited the amusement park. 56% of the visitors were adults, 16% were teenagers, and 28% were children

ages 12 and under. Find the number of adults, teenagers, and children that visited the part.
Mathematics
1 answer:
liberstina [14]3 years ago
3 0

Answer:

Adults

= Percentage of adults * Total number of people who visited

= 56% * 1,575

= 882 adults

Teenagers

= Percentage of teenagers * Total number of people who visited

= 16% * 1,575

= 252 teenagers

Children

= 28% * 1,575

= 441 children

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castortr0y [4]

<u>angle B and angle G</u> are congruent because of the alternate exterior angles theorem.

<u>angle A and angle H</u> are also congruent, because of the same theorem.

So, the answers are <u>A, and D</u>


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I'LL GIVE BRAINLIEST !!! FASTER<br><br>please explain how do you get the answer !​
aliya0001 [1]

Answer:

70

Step-by-step explanation:

we have the angle of vertex in the isosceles triangle = 180-2*bottom coner= 180-65/2=50

3 angles in the equilateral triangle are equal to 60

we have 50 + 60 +h =the angle of PQR =180

h=70

7 0
3 years ago
Read 2 more answers
Which of the following points lies on the graph of y = 3x-5
andreyandreev [35.5K]

Answer:

(4,7)

Step-by-step explanation:

I have attached what the graph looks like below.

Looking at it, you will see the point (4,7) lies on the graph.

So the answer is 2)

6 0
3 years ago
The grocery store is mile from Katie’s house and the gas station is 0.5 mile from Katie’s house.
Effectus [21]

Answer:

the answer would be a

Step-by-step explanation

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8 0
3 years ago
A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.
Serjik [45]

Answer:

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

t=\sqrt{\frac{2h}{a}} ,where h=height of release

a=acceleration

a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}

Where I=moment of inertia

a for hoop

a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}

a=\frac{g\sin \theta }{2}

a for Uniform solid cylinder

a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}

a=\frac{2g\sin \theta }{3}

a for spherical shell

a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}

a=\frac{3g\sin \theta }{5}

a for Uniform Solid

a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}

a=\frac{5g\sin \theta }{7}

time taken will be inversely proportional to the square root of acceleration

t_1=k\sqrt{2}=1.414k

t_2=k\sqrt{\frac{3}{2}}=1.224k

t_3=k\sqrt{\frac{5}{3}}=1.2909k

t_4=k\sqrt{\frac{7}{5}}=1.183k

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0
3 years ago
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