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1 year ago

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Marcel plugged in his work tablet and phone. the phone had a battery charge of 13% and started increasing by 2 percentage points

every 3 minutes. the tablet had charge of 25% and started increasing by 1 percentage point every 3 minutes.
Let t represent the time, in minutes, since Marcel plugged in the phone and tablet.

Complete the inequality to represent the times when the phone would have at least as much battery charge as the tablet

2 answers:

Verdich [7]1 year ago

7 0

Answer:

t ≥ 36

Step-by-step explanation:

took a while but i had to try a lot of the numbers. hope it helped, mark me brainleist.

ch4aika [34]1 year ago

5 0

The inequality to represent the times when the phone would have at least as much battery charge as the tablet is <u>t ≥ 36</u>.

We know that t is the time, in minutes, since Marcel plugged in the phone and the tablet.

<u>For phone</u>:-

Initial battery charge = 13%.

Rate of charging = 2% points in 3 minutes = 2/3 % in 1 minute.

Thus, the total charge on the phone after t minutes = 13% + t(2/3)%.

<u>For tablet</u>:-

Initial battery charge = 25%.

Rate of charging = 1% points in 3 minutes = 1/3 % in 1 minute.

Thus, the total charge on the tablet after t minutes = 25% + t(1/3)%.

We are asked to complete the inequality to represent the times when the phone would have at least as much battery charge as the tablet.

Since the phone is required to have at least as much battery charge as the tablet, we can show this as:

The total charge on the phone after t minutes ≥ The total charge on the tablet after t minutes,

or, 13% + t(2/3)% ≥ 25% + t(1/3)%,

or, t(2/3)% - t(1/3)% ≥ 25% - 13%,

or, t(1/3)% ≥ 12%,

or, t ≥ 12%/(1/3)%,

or, t ≥ 36.

Thus, the inequality to represent the times when the phone would have at least as much battery charge as the tablet is <u>t ≥ 36</u>.

Learn more about inequalities at

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2 years ago

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3 years ago

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<u>The formula for Mean deviation</u> $= \sum \frac{|x_{i}- \mu|}{N}$ , where $x_{i}$ is the given data from $i=1$ to $i=10$ , $\mu$ is the mean of the data and $N$ is the total number of data. So.....

$|x_{1}-\mu |= |34-33.9|=0.1\\ |x_{2}-\mu|=|35-33.9|=1.1\\ |x_{3}-\mu |=|41-33.9|=7.1\\ |x_{4}-\mu |=|28-33.9|=5.9\\ |x_{5}-\mu |=|26-33.9|=7.9\\ |x_{6}-\mu |=|29-33.9|=4.9\\|x_{7}-\mu |=|32-33.9|=1.9\\ |x_{8}-\mu |=|36-33.9|=2.1\\ |x_{9}-\mu |=|38-33.9|=4.1\\ |x_{10}-\mu |=|40-33.9|=6.1$

So, the Mean deviation $=\sum \frac{|x_{i}-\mu |}{N}=\frac{0.1+1.1+7.1+5.9+7.9+4.9+1.9+2.1+4.1+6.1}{10} = \frac{41.2}{10}=4.12$

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3 years ago