To answer this you can use a guess and check strategy.

Start with 5 minutes using it (25% lost) and 10 minutes charging (20% gained).

Try 7 minutes using it (35% lost) and 14 minutes charging (28%).

Next, 9 minutes (45%) and 18 minutes(36%).

10 minutes (50%) and 20 minutes (40%). This is the difference.

She used it for 10 minutes and charged it for 20 minutes.

6 0

3 years ago

Read 2 more answers

Use the graph to find the following.

Levart [38]

a) Local maximum values of g are: 0 and 3

b) The values at which g has local maximum are : -2 and 2

<h3>What is a maximum point?</h3>

A maximum point is the point at which the gradient of the curve changes from positive to negative.

Analysis:

The inverted v-curves all have maximum values and they are two.

The one at x = -2 and the one at x = 2 for in both cases their maximum values are y = 0 and y = 3 respectively.

Learn more about minimum and maximum points: brainly.com/question/14993153

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5 0

2 years ago

The body temperatures of adults are normally distributed with a mean of 98.6degrees° F and a standard deviation of 0.60degrees°

Schach [20]

Answer:

97.72% probability that their mean body temperature is greater than 98.4degrees° F.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$