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s344n2d4d5 [400]
4 years ago
14

Paige spent 1/5 of an hour playing on her phone. That used up 1/5 of her battery. How long would she have to play on her phone t

o yse the entire battery
Mathematics
1 answer:
OlgaM077 [116]4 years ago
5 0

Answer: An hour

Step-by-step explanation:

Since Piage Spent 1/5 of an hour to use 1/5 of her battery.

If she spends 5/5 of an hour which is 1hour, she will use up 5/5 of her battery, which is 1battery.

Thanks. I hope this is helpful?

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If I were you I will think the answer is B
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Write these as normal numbers.
ELEN [110]

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a) 510000000 c) 24.4 d) 502.3

Step-by-step explanation:

7 0
3 years ago
In the xy-plane, the equations x + 2y = 10 and
konstantin123 [22]

By solving given equations, the value of c is 30.

Given two equations

x + 2y = 10 and

3x + 6y = c

These lines represents the same line for some constant c.

Value of c:

x + 2y = 10-------------(1)

3x + 6y = c-------------(2)

Dividing equation (2) by 3

\frac{3x}{3}+ \frac{6y}{3} =\frac{c}{3}

After solving the above equation, we get

x + 2y = c/3-----------(3)

Remember that a line is written as ax + by = c, in our case, both lines have a =1 and b = 2. Therefore, in orther that the two lines are equal, we need that, 10 = c/3

c = 10 × 3 = 30

c = 30

Therefore,

The value of c is 30.

Find out more information about equations here

brainly.com/question/21952977

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7 0
2 years ago
You have a six-sided die that you roll once. Let Ri denote the event that the roll is i. Let G j denote the event that the roll
Misha Larkins [42]

Answer:

a. P (R3 | G1)=\frac{1}{5}

b. P (R6| G3)= \frac{1}{3}

c. P(G3|E)=\frac{2}{3}

d. P (E|G3)=\frac{2}{3}

Step-by-step explanation:

The sample space associated with the random experiment of throwing a dice is is the equiprobable space {R1, R2, R3, R4, R5, R6}. Then,

a. The conditional probability that 3 is rolled given that the roll is greater than 1? P (R3 | G1) = \frac{P (R3\bigcap G1)}{P(G1)} = \frac{1/6}{5/6} = \frac{1}{5}

b. What is the conditional probability that 6 is rolled given that the roll is greater than 3? P (R6| G3) = \frac{P (R6\bigcap G3)}{P(G3)} = \frac{1/6}{3/6} = \frac{1}{3}

c. What is P [GIE], the conditional probability that the roll is greater than 3 given that the roll is even? P(G3|E) = \frac{P (G3\bigcap E)}{P(E)} = \frac{2/6}{3/6} = \frac{2}{3}

d. Given that the roll is greater than 3, what is the conditional probability that the roll is even? P (E|G3) = \frac{P (E\bigcap G3)}{P(G3)} = \frac{2/6}{3/6} = \frac{2}{3}

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3 years ago
A quadrilateral has no pairs of parallel sides. what two shapes could it be
Svetach [21]
A trapezium I believe
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3 years ago
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