M2 -5m - 24 pls answer my questions<br>

The option that is true with regard to the following functions is Option B. "The domain g(x) and h(x) include all real number while the domain of i(x) and h(x) are restricted"

<h3>What is the explanation for the above?</h3>

Lets examine f(x) = 3x + 14

Note that this function is indicative of a straight-line. See the attached graph for function 1. Note that it doesn't have any end points. That is, it is Asymptote.

Let us examine h(x) = 3ˣ + 1

This represents an exponential graph. Just like the function above it doesn't have any end point. It however has an asymptote:

y = 0

Let us look at F'(x) = Log₃ (x = 1).

This is indicative of logarithm graph. It doesn't have any end but has an asymptote x == 0

Let us take a look at g(x) = X⁴ + 3x² - 14

Notice that in the mid point there is an end point given as (0, -14). Thus, it is correct to state that the function in Option B is the only one that exhibits end behavior and as such is restricted.

Learn more about functions:
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1 year ago

The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of ten batterie

Fittoniya [83]

Answer:

The confidence interval is $25.16 < \mu < 26.85$

Step-by-step explanation:

From the question we are given a data set

25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7.

The mean of the this sample data is

$\= x = \frac{\sum x_i}{n}$

where is the sample size with values n = 10

$\= x = \frac{25.5+ 26.1+ 26.8+23.2+ 24.2+ 28.4+ 25.0+ 27.8+ 27.3+ 25.7}{10}$

$\= x = 26$

The standard deviation is evaluated as

$\sigma = \sqrt{\frac{\sum (x-\= x)}{n} }$

substituting values

$= \sqrt{\frac{ ( 25.5-26)^2, (26.1-26)^2, (26.8-26)^2, (23.2-26)^2}{10} }$

$\cdot \ \cdot \ \cdot \sqrt{\frac{ ( 24.2-26)^2, (28.4-26)^2+( 25.0-26)^2+ (27.8-26)^2+( 27.3-26)^2+( 25.7-26)^2}{10} }$

$\sigma = 1.625$

The confidence level is given as 90% hence the level of significance is calculated as

$\alpha = 100 -90$

$\alpha =10$ %

$\alpha = 0.10$

Now the critical values of $\frac{\alpha }{2}$ is obtained from the normal distribution table as

$Z_{\frac{\alpha }{2} } = 1.645$

The reason we are obtaining the critical values of $\frac{\alpha }{2}$ instead of $\alpha$ is because we are considering two tails of the area under the normal curve