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3 years ago

Given h(x) = 2x – 5, find h(6).

Mathematics

2 answers:

pantera1 [17]3 years ago

6 0

Answer:

Step-by-step explanation:

h(X)=2x-5

h(6)=?

we have to put 6 in place of x ,

h(6)=2*6- 5

h(6)=12-5

so,h(6)=7

Ludmilka [50]3 years ago

3 0

Answer:

Step-by-step explanation:

h(6)=2(6)–5

h(6)=12–5

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andreyandreev [35.5K]

Answer:

24/35

Step-by-step explanation:

1) Convert your mixed number (1 and 1/7) to an improper fraction by multiplying your whole number (1) and your denominator (7) which gives you 7. Then add your numerator (1) to that which gets you 8. Then take the original denominator (7) and turn it into improper where your new numerator is 8 and your denominator stays 7. (If that doesn't make much sense then you can look up how to change mixed numbers into improper fractions.)

2) Multiply your numbers straight across.

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3 years ago

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3 years ago

A manufacturer of electronic kits has found that the mean time required for novices to assemble its new circuit tester is 3 hour

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At the null hypothesis, we test if there has been no improvement, that is, the time is still of 3 hours. Hence:

$H_0: \mu = 3$

At the alternative hypothesis, we test if there has been improvement, that is, the time is of less than 3 hours. Thus:

$H_1: \mu < 3$

We have the standard deviation for the population, thus, the z-distribution is used. The test statistic is given by:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.
$\mu$ is the value tested at the null hypothesis.
$\sigma$ is the standard deviation of the population.
n is the sample size.

In this problem, the values of the parameters are: $\mu = 3, \sigma = 0.2, \overline{x} = 2.9, n = 15$ .

Then, the value of the test statistic is:

$z = \frac{2.9 - 3}{\frac{0.2}{\sqrt{15}}}$

$z = -1.94$

The p-value of the test is the probability of finding a sample mean of 2.9 hours or less, which is the p-value of z = -1.94.

Looking at the z-table, z = -1.94 has a p-value of 0.0262.

Since the p-value of the test is of 0.0262 < 0.05, it can be concluded that the new booklet is effective.

A similar problem is given at brainly.com/question/25413422

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