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zloy xaker [14]
3 years ago
15

use the graph of y=e^x to evaluate the expression e^0. round the solution to the nearest tenth if necessary​

Mathematics
1 answer:
Wittaler [7]3 years ago
7 0

In the attached picture you have the graph of y=e^x. To evaluate the function at x=0, you'll have to start from 0 on the x axis, and travel vertically until you meet the graph.

In this case, if you start from the origin, you have to go upwards one unit, until you meet the graph at (0,1).

So, if x=0, y=1.

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2 apples and 1 mango

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4 years ago
A​ 9-year-old girl did a science fair experiment in which she tested professional touch therapists to see if they could sense he
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We conclude that the the touch therapists does not use a method equivalent to random guesses.

Yes, the results suggest that touch therapists are​ effective.

Step-by-step explanation:

We are given that in a 9-year-old girl did a science fair experiment in which she tested professional touch therapists to see if they could sense her energy field.

Among 264264 ​trials, the touch therapists were correct 105105 times.

<u><em>Let p = proportion that touch therapists uses a random guess method.</em></u>

Here, random guess means; p = 50%

So, Null Hypothesis, H_0 : p = 50%     {means that the touch therapists use a method equivalent to random guesses}

Alternate Hypothesis, H_A : p \neq 50%     {means that the touch therapists does not use a method equivalent to random guesses}

The test statistics that would be used here <u>One-sample z proportion statistics</u>;

                         T.S. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p  = sample proportion of touch therapists were correct = \frac{105}{264} = 0.39

            n = sample of trials = 264

So, <em><u>test statistics</u></em>  =  \frac{\frac{105}{264} -0.50}{\sqrt{\frac{\frac{105}{264}(1-\frac{105}{264})}{264} } }  

                                =  -3.395

The value of z test statistics is -3.395.

<em>Now, at 0.05 significance level the z table gives critical value of -1.96 and 1.96 for two-tailed test. Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>

Therefore, we conclude that the the touch therapists does not use a method equivalent to random guesses.

Yes, the results suggest that touch therapists are​ effective.

3 0
3 years ago
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