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Sergio [31]
3 years ago
12

What is the FDP of 0.3

Mathematics
1 answer:
exis [7]3 years ago
8 0

Answer:

  0.3 = 3/10 = 30%

Step-by-step explanation:

The reading of the number as "three tenths" tells you how to write the fraction:

  0.3 = 3/10

You can also express the same value in hundredths. Of course /100 is just a long way to write %, so ...

  0.3 = 0.30 = 30/100 = 30%

0.3 = 3/10 = 30%

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William bought some tickets to see his favorite singer. He bought some adult tickets and some children tickets, for a total of 1
Viefleur [7K]
Hello!

Let 'a' stand for adult tickets and 'c' for children tickets.

a + c = 15

(Both adult and children tickets make up a total of 15 tickets).

30a + 20c = 330

a = 3

c = 12

A N S W E R:

William bought 3 adults tickets and 12 children's tickets.

5 0
3 years ago
Plz plz help test due soon
Dvinal [7]

Answer:

22

Step-by-step explanation:

8 0
3 years ago
What number should be added to 1253 to get 3000​
grandymaker [24]
1847..............................
4 0
3 years ago
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A homeowner has an octagonal gazebo inside a circular area. Each vertex of the gazebo lies on the circumference of the circular
Tcecarenko [31]

If we draw the diagonals of the octagonal gazebo, the 4 diagonals divide the octagon into 8 triangles.

Note that each triangle is an isosceles triangle whose equal sides are x, the radius of the circle.

The top angle of each triangle is obtained by dividing the full angle by 8.

So, each top angle = \frac{360}{8}

= 45°

Now, in fig., consider one of the triangles Δ OAB. Draw an altitude OC from O to the opposite side AB.

This altitude OC bisects the top angle 45°.

Therefore, ∠ AOC = 22.5°.

Now, in Δ AOC,

sin 22.5=\frac{AC}{OA}

=\frac{AC}{x}

So, AC = x sin 22.5°

Note that, AB = 2 AC.

Therefore, AB = 2x sin 22.5°.

Also, cos 22.5=\frac{OC}{OA}

=\frac{OC}{x}

So, OC = x cos 22.5°.

Area of Δ AOB = \frac{1}{2}(AB)(OC)

= \frac{1}{2} × (2x sin 22.5°) × (x cos 22.5°)

= \frac{1}{2} x^{2} (2 sin 22.5° cos 22.5°)

= \frac{1}{2} x^{2} sin 45°

= x^{2} / 2\sqrt{2}

Area of the octagonal gazebo = 8 × one triangular area

= 8 × (x^{2} / 2\sqrt{2})

=2\sqrt{2} x^{2}

=2.828x^{2}

Area required for mulch = circular area - area of the gazebo

=3.14x^{2} -2.828x^{2}

=0.312x^{2}

Now, cost per unit area = $1.50.

Hence, total cost g(m) = area × cost per unit area

Total cost g(m) = 0.312x^{2} × 1.5

=0.468x^{2}

Hence, total cost g(m) = 0.468x^{2}.

4 0
3 years ago
I NEED HELP PLEASE <br> ITS DUE SOOM
tekilochka [14]

Answer:

Tyler is correct. The temperature dropped at a rate of about 4° per hour between 4 and 6, while the temperature dropped at about 2.25° per hour between 6 and 10.

Edit: Explanation

The question is asking about which window of time had a <em>faster</em> decline in temperature, not a larger total change in temperature.

In a 2 hour timeframe, the temperature dropped 8°. (4-6 PM)

In a separate 4 hour timeframe, the temperature dropped 9°. (6-10 PM)

To find which window had a faster change in temp, I took the total temperature drop for each timeframe, then divided it by the number of hours each drop took.

8° / 2 = 4° per hour for 4-6 PM

9° / 4 = 2.25° per hour from 6-10 PM

Since the speed at which the temperature dropped per hour was greater from 4-6 PM than 6-10 PM, Tyler was correct.

6 0
3 years ago
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