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4 years ago

Must a function that is decreasing over a given interval always be negative over that same interval? Explain.

Mathematics

2 answers:

denis23 [38]4 years ago

6 0

Answer:

For a function to be decreasing over an interval, the outputs are getting smaller as the inputs of the function are getting larger.

The outputs of a decreasing interval could be positive or negative.

For a function to be negative over an interval, the outputs must be negative, while the inputs could be positive or negative.

Step-by-step explanation:

kompoz [17]4 years ago

4 0

No. The sign of a function is not related with the fact that it is growing of decreasing.

If the function in the decreasing in the interval (a,b) it means that f(b) < f(a), but yet both f(b) and f(a) may be positive.

For example imagin the function y = - x. It is a straight line and it is decreasing for any interval that you take; this is (-∞, ∞).

If you take the interval (- ∞, 0) the function is decreasing but the values are positive. If you take the interval (0, ∞) the function is decreasing and the values are negative.

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Let f(x)=(x+3)^2and g(x)=f(x)-7. Which of the following shows g(x)?

Mila [183]

F(x) = (x + 3)²

g(x) = f(x) - 7.

g(x) = (x + 3)² - 7

g(x) = (x + 3)(x + 3) - 7

g(x) = x*x + x*3 +3*x + 3*3 - 7

g(x) = x² + 3x + 3x + 9 - 7

g(x) = x² + 6x + 2

6 0

3 years ago

Which expression is equivalent to

Mila [183]

Answer:

Yhe answer is C.

Step-by-step explanation:

First, you have to get rids of brackets by multiplying :

$10( - \frac{4}{5} x + 3) - 2x$

$= 10( - \frac{4}{5} x) + 10(3) - 2x$

$= - 8x + 30 - 2x$

Next you have to collect the like-terms :

$- 8x + 30 - 2x$

$= - 10x + 30$

8 0

3 years ago

Does this represent two quantities that are proportional

Nadya [2.5K]

Answer:

from looking at this we can see

1*4=4

2*4=8

3*4=12

6*4=24

Hope This Helps!!!

3 0

2 years ago

Im bad at number lines oh

liq [111]

-2 1/3 would be in between -2 and -3, but closer to -2.

4 0

3 years ago

Read 2 more answers

Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst 1, resulting in an average yi

aalyn [17]

Answer:

a) $t=\frac{(91-85)-(0)}{2.490\sqrt{\frac{1}{12}}+\frac{1}{15}}=6.222$

$df=12+15-2=25$

$p_v =P(t_{25}>6.222) =8.26x10^{-7}$

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

b) $(91-85) -2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =3.309$

$(91-85) +2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =8.691$

Step-by-step explanation:

Notation and hypothesis

When we have two independent samples from two normal distributions with equal variances we are assuming that

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}}+\frac{1}{n_2}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

This last one is an unbiased estimator of the common variance $\sigma^2$

Part a

The system of hypothesis on this case are:

Null hypothesis: $\mu_2 \leq \mu_1$

Alternative hypothesis: $\mu_2 > \mu_1$

Or equivalently:

Null hypothesis: $\mu_2 - \mu_1 \leq 0$

Alternative hypothesis: $\mu_2 -\mu_1 > 0$

Our notation on this case :

$n_1 =12$ represent the sample size for group 1

$n_2 =15$ represent the sample size for group 2

$\bar X_1 =85$ represent the sample mean for the group 1

$\bar X_2 =91$ represent the sample mean for the group 2

$s_1=3$ represent the sample standard deviation for group 1

$s_2=2$ represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

$\S^2_p =\frac{(12-1)(3)^2 +(15 -1)(2)^2}{12 +15 -2}=6.2$

And the deviation would be just the square root of the variance:

$S_p=2.490$

Calculate the statistic

And now we can calculate the statistic:

$t=\frac{(91-85)-(0)}{2.490\sqrt{\frac{1}{12}}+\frac{1}{15}}=6.222$

Now we can calculate the degrees of freedom given by:

$df=12+15-2=25$

Calculate the p value

And now we can calculate the p value using the altenative hypothesis:

$p_v =P(t_{25}>6.222) =8.26x10^{-7}$

Conclusion

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

Part b

For this case the confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} S_p \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

For the 99% of confidence we have $\alpha=1-0.99 = 0.01$ and $\alpha/2 =0.005$ and the critical value with 25 degrees of freedom on the t distribution is $t_{\alpha/2}= 2.79$

And replacing we got:

$(91-85) -2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =3.309$

$(91-85) +2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =8.691$

7 0

3 years ago