Explain how you know without computing whether the quotient of 1/2 divided by 6 is greater than or less than 1

1 answer:

6 0

It is less than 1, because 1/2 is also .5, so that is already less than 1 whole.
Also, any number under 6 divided by 6 (for example: 5 divided by 6, 3 divided by 6) has to be less than one, since 6 can't evenly go into a number lower than it.

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The concentration of hexane (a common solvent) was measured in units of micrograms per liter for a simple random sample of sixte

goldenfox [79]

Answer:

Yes, it can be concluded that the mean hexane concentration is less in treated water than in unsaturated water

Step-by-step explanation:

The number of of specimen in the samples of untreated water, n₁ = 16

The sample mean, $\overline x_1$ = 228.0

The sample standard deviation, s₁ = 4.3

The number of of specimen in the samples of treated water, n₂ = 20

The sample mean, $\overline x_2$ = 224.6

The sample standard deviation, s₂ = 5.0

The level of significance = 0.10

The null hypothesis, H₀; $\overline x_1$ ≥ $\overline x_2$

The alternative hypothesis, Hₐ; $\overline x_1$ < $\overline x_2$

The degrees of freedom = 16 - 1 = 15

The test statistic, $t_{\alpha}$ = 1.341

$t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}-\dfrac{s _{2}^{2}}{n_{2}}}}$

Plugging in the values, we get;

$t=\dfrac{(224.6- 228.0)}{\sqrt{\dfrac{5.0^{2}}{20} -\dfrac{4.3^{2} }{16}}} \approx -11.0675$

Given that the t-value is large, the corresponding p-value is low, therefore, we fail to reject the null hypothesis and there is considerable statistical evidence to suggest that the mean hexane concentration is less in treated than in untreated water, therefore, we have; $\overline x_1$ ≥ $\overline x_2$

6 0

3 years ago

What is the value of f(3) when<br> f(3) when f(x) = 4x+1

IgorLugansk [536]

Answer:

f(3) = 13

Step-by-step explanation:

f(x) = 4x+1

f(3) = 4(3) + 1

f(3) = 12 + 1

f(3) = 13

5 0