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Ratling [72]
3 years ago
6

If start fraction 1 over 3 end fraction is equivalent to 33start fraction 1 over 3 end fraction%, what percent is equivalent to

two-thirds? A. 33two-thirds% B. 150% C. 66two-thirds% D. 65%
Mathematics
1 answer:
kotegsom [21]3 years ago
5 0

Answer:

The answer is C 66%

Step-by-step explanation:

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Please ASAP.
Flauer [41]

Answer:

2/3

Step-by-step explanation:

42/63 = 2/3

15/22.5 = 2/3

8 0
3 years ago
Read 2 more answers
Customers of a phone company can choose between two service plans for long distance calls. The first plan has a $24 monthly fee
JulsSmile [24]

Answer:

600 minutes

Step-by-step explanation:

If we write both situations as an equation, we get:

y1 = 24 + 0.15x

<em>y1 </em><em>:</em><em> </em><em>total </em><em>cost </em><em>paid </em><em>in </em><em>first </em><em>plan</em>

<em>x </em><em>:</em><em> </em><em>total minutes </em><em>of </em><em>calls</em>

y2 = 0.19x

<em>y2 </em><em>:</em><em> </em><em>total </em><em>cost </em><em>in </em><em>second </em><em>plan</em>

<em>x:</em><em> </em><em>total </em><em>min</em><em>utes </em><em>of </em><em>call</em>

We are now looking for the situation where the total cost in the two plans is equal, so

y1 = y2

this gives

24 + 0.15x = 0.19x

<=> 0.04x = 24

<=> x = 600

7 0
3 years ago
I will mark brainliest !
harina [27]

Answer:

90 degrees

Step-by-step explanation:

A circle circumscribed over a right triangle is the same as a circle circumscribed over a rectangle built from the extension of the right triangle.

That tells us that the hypotenuse is the diameter of the circle in that case.

Here, we can see a triangle that has the hypotenuse as the diameter of the circumscribed circle. We can infer that the triangle is a right triangle.

4 0
3 years ago
J^2 over 2j times 2j over 3g
Serhud [2]

Answer:

D

Step-by-step explanation:

given \frac{j^2}{2j} × \frac{2j}{3g}

the 2j on the denominator of the first fraction cancels with the 2j on the numerator of the second fraction, leaving the expression in simplified form as

= \frac{j^2}{3g} → D


5 0
3 years ago
The population of a town has approximately doubled every 17 years since 1950. If the equation P=Po2k, where Po is the population
natka813 [3]

The population of a town has approximately doubled every 17 years since 1950.

the equation P=P_02^k where Po is the population of the town in 1950, is used to model the population, P, of the town t years after 1950.

When t=17 yrs

   P=2p_{0}      

for 1 year

The equation becomes

P = P_02^\frac{t}{17}                               -------------(1)

Our original equation is

P=P_02^k----------------------------------(2)

equating expression 1  and 2

P_02^\frac{t}{17}=P_{0}2^k

Cancelling P_{0}  from both sides  we get

2^\frac{t}{17}=2^{k}

t/17=k

⇒k=t/17 is the solution.






4 0
3 years ago
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