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3 years ago

Translate the algebraic expression

Mathematics

2 answers:

Radda [10]3 years ago

7 0

Answer:

The quotient of –16 and a number

Negative 16 divided by a number

Step-by-step explanation:

ki77a [65]3 years ago

3 0

Answer:

The quotient of –16 and a number

Negative 16 divided by a number

Step-by-step explanation:

We are required to translate the algebraic expression -16/n into words. Before we translate, it will be necessary to note that quotient is the outcome or result of division of a term by another term.

For example, if the expression that we are faced with is 20/5, then the quotient of 20 and 5 is 4.

If we are required to solve 12/2 then 6 which is the result is the quotient of 12 and 2.

Now, if we are presented with the algebraic expression:

-16/n

Then, there are two ways we can successfully translate this expression and they are:

a. Negative sixteen divided by a number.

b. The quotient of negative sixteen and a number.

Let us consider the options given to us:-

*** The difference of -16 and a number ---------- this is wrong because the expression doesn't entail subtracting a number from negative sixteen or vice versa

***The quotient of -16 and a number ---------- This is correct.

*** A number divided by -16 -------------- This is wrong. on the contrary, -16 is being divided by a number.

*** Negative 16 divided by a number --------- This is correct

*** The quotient of a number and 16 -------- This is wrong because one of the figures is negative 16 and not +16.

The two options that successfully translated the algebraic expression are options B and D.

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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; $\overline x_1$ ≥ $\overline x_2$

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GuDViN [60]

Answer:

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Step-by-step explanation:

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