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3 years ago

9

Dans computer reads 68% of battery left after 1 hour. If the rate of using power remains the same, about how much longer will hi

s battery provide power?

1 answer:

Tanya [424]3 years ago

7 0

Answer:

28 minutes

Step-by-step explanation:

Dans computer reads 68% of battery left after 1 hour. This means that 1 hour is 68% of the computer's entire battery life.

Let the entire battery life be x (in hours). Hence:

$\frac{1}{x} * 100 = 68$

Therefore:

$\frac{100}{x} = 68\\\\x = \frac{100}{68}$

=> x = 1.47 hrs = 1 hr 28 minutes

Hence, the entire battery life is 1 hr 28 minutes.

If 1 hour has elapsed, the remaining time left for his battery to provide power will therefore be:

1 hr 28 minutes - 1 hr = 28 minutes

His battery will provide power for 28 more minutes.

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A restaurant charges $5.00 for a slice of pizza that is 1/3 of a pizza and $4.50 for a slice that is 1/5 of a pizza. One day the

katrin [286]

Answer:

<h3>5ths</h3>

Step-by-step explanation:

Given that:

Price of whole pizza (A) = 5 × 3 = $15
Price of whole pizza (B) = 4.50 × 5 = $22.50

They make 8 whole pizzas.

Therefore

Thirds (A) = 15 × 8 = $120
Fifths (B) = 22.50 × 8 = $180

They'll make more money if they slice it into fifths.

180 - 120 = $60

Because if they slice it into thirds, total profit will be $60 less

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2 years ago

7. The hexagon GIKMPR is regular. The dashed line segments form 30 degree angles. what is the image of oh after a rotation of 18

velikii [3]

Answer:

The image of OH is ON.

Step-by-step explanation:

Te figure GIKMPR is a regular hexagon. The number of vertices of a regular hexagon is 6. The central angle between any two consecutive vertices is 60 degree.

The dashed line segments form 30 degree angles.

If we rotate the hexagon 180 degree about O, then the each point shifts to 6th place from its original place.

Since we rotate the hexagon 180 degree about O, so the image and preimage lies on a straight line. Because a straight line make angle of 180 degree.

The line OH and ON lies on a straight line therefore the image of OH is ON.

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3 years ago

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3 years ago

Simplify −2xy + 3x − 2xy + 3x <br><br> A. 4xy +6x B. 2xy C. −4xy + 6x D. 4xy − 6x

marysya [2.9K]

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3 years ago

Read 2 more answers

Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodo

Shkiper50 [21]

Answer:

a) $P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.6288$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.1954=0.4335$

b) $P(4\leq X\leq 8)=0.1954+0.1563+0.1042+0.0595+0.0298=0.5452$

c) $P(X \geq 8) = 1-P(X$

d) $P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

Step-by-step explanation:

Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that $X \sim Poisson(\lambda=4)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda=4$ , $Var(X)=\lambda=2$ , $Sd(X)=2$

a. Compute both P(X≤4) and P(X<4).

$P(X\leq 4)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)+P(X=4)$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-4} 4^0}{0!}=e^{-4}=0.0183$

$P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733$

$P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465$

$P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.9646$

$P(X< 4)=P(X\leq 3)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.5311=0.7692$

b. Compute P(4≤X≤ 8).

$P(4\leq X\leq 8)=P(X=4)+P(X=5)+ P(X=6)+P(X=7)+P(X=8)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(X=7)=\frac{e^{-4} 4^7}{7!}=0.0595$

$P(X=8)=\frac{e^{-4} 4^8}{8!}=0.0298$

$P(4\leq X\leq 8)=0.1954+0.1563+ 0.1042+0.0595+0.0298=0.5452$

c. Compute P(8≤ X).

$P(X \geq 8) = 1-P(X$

$P(X \geq 8) = 1-P(X$

d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?

The mean is 4 and the deviation is 2, so we want this probability

$P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

4 0

3 years ago