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3 years ago

Solve the quadratic equation by completing the square x²+12x+26=0

Mathematics

1 answer:

vova2212 [387]3 years ago

4 0

Answer:

x={-13,-2}

Step-by-step explanation:

(x+13)(x+2)=0

x+13=0

x=-13

x+2=0

x=-2

x={-13,-2}

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4= Sx+10/2<br><br> that’s a fraction btw

frozen [14]

Answer:

4= Sx+10/2 simplifies to -s*x-1=0

Step-by-step explanation:

8 0

3 years ago

Please help I don’t understand number 16

Allisa [31]

Let's go step by step:

a) You are given four couples of x and y values which model the relationship between the number of drinks and their cost. So, the couple $(x,y)=(0,0)$ means that if you buy zero drinks, you spend no money. That makes sense. The next information we have is $(x,y) = (2,3)$ , which means that two drinks cost 3 dollars, and so on.

So, you simply need to draw on the grid the four points

$(0,0),\ (2,3),\ (4,6),\ (6,9)$

b) The domain is the set of inputs. Since we only know the value of the function on four different points (we know the price for 0,2,4 and 6 drinks), the domain is discrete. In fact, a continuous domain must contain an interval (for example, $[1,2]$ is a continuous domain), whereas if you pick a certain number of points (like in this case: we picked 0,2,4 and 6), the domain is discrete.

c) Once the points are drawn on the grid, you can see that they all lie on the same line. To find that line, we will only need two of those points (once two points are fixed, there is only one line passing through them). In general, the equation of the line passing through $P = (P_x,P_y)$ and $Q = (Q_x,Q_y)$ is

$\cfrac{x-P_x}{Q_x-P_x} = \cfrac{y-P_y}{Q_y-P_y}$

Let's choose, for example, the first two points. The equation is

$\cfrac{x-0}{2-0} = \cfrac{y-0}{3-0} \iff \cfrac{x}{2} = \cfrac{y}{3} \iff y=\cfrac{3}{2}x$

d) Now that we know the equation of the line, we can compute the cost of any number of drinks: the equation of the line is a function that associates a cost, y, with every possible number of drinks, x.

Of course, some associations will be odd - we can compute the cost of $\sqrt{2}$ drink, but what would it mean?

Anyway, the question about the cost of two drinks seems more than reasonable, so let's see which y value is associate with the particular x value of 2:

$y = \cfrac{3}{2} x \implies y = \cfrac{3}{2}\cdot 2 = 3$

So, two drinks cost 3 dollars.

3 0

3 years ago

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 $t_{\alpha/2}= 2.79$