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2 years ago

HELP!!! What’s the domain and range? PLS!!!!

Mathematics

1 answer:

Arte-miy333 [17]2 years ago

7 0

Answer:

A) Vertex: (3,1) Maximum, Domain: all real numbers, Range: y is less than or equal to 1.

B) Vertex: (1,-5) Minimum, Domain: all real numbers, Range: y is greater than or equal to -5

C) Vertex: (0,1) Minimum, Domain: all real numbers, Range: y is greater than or equal to 1

D) Vertex: (4,0) Maximum, Domain: all real numbers, Range: y is less than or equal to 0.

E) Vertex: (4,1) Minimum, Domain: all real numbers, Range: y is greater than or equal to 1

F) Vertex: (-1,4) Maximum, Domain: all real numbers, Range: y is less than or equal to 4.

G) Vertex: (2,-4) Minimum, Domain: all real numbers, Range: y is greater than or equal to -4

H) Vertex: (-4,-3) Minimum, Domain: all real numbers, Range: y is greater than or equal to -3

I) Vertex: (2,3) Maximum, Domain: all real numbers, Range: y is less than or equal to 3.

Step-by-step explanation:

Sorry it took forever to type

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Answer:

y-5=1/4(x-3)

Step-by-step explanation:

first you need to find the slope.

5-7 over 3-6 = -1/-4 or 1/4

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A toy rocket is launched from a platform that is 48 ft high. The rocket height above the ground is modeled by h=-16t^2+32t+48. a

Komok [63]

Answer:

a) 64 feet

b) 3 seconds

Step-by-step explanation:

The maximum height of $h=h(t)$ can be bound by finding the y-coordinate of the vertex of $y=-16x^2+32x+48$ .

Compare this equation to $y=ax^2+bx+c$ to find the values of $a,b,\text{ and } c$ .

$a=-16$

$b=32$

$c=48$ .

The x-coordinate of the vertex can be found by evaluating:

$\frac{-b}{2a}=\frac{-32}{2(-16)$

$\frac{-b}{2a}=\frac{-32}{-32}$

$\frac{-b}{2a}=1$

So the x-coordinate of the vertex is 1.

The y-coordinate can be found be evaluating $y=-16x^2+32x+48$ at $x=1$ :

$y=-16(1)^2+32(1)+48$

$y=-16+32+48$

$y=16+48$

$y=64$

So the maximum height of the rocket is 64 ft high.

When the rocket hit's the ground the height that the rocket will be from the ground is 0 ft.

So we are trying to find the second t such that:

$0=-16t^2+32t+48$

I'm going to divide both sides by -16:

$0=t^2-2t-3$

Now we need to find two numbers that multiply to be -3 and add to be -2.

Those numbers are -3 and 1 since (-3)(1)=-3 and (-3)+(1)=-2.

$0=(t-3)(t+1)$

This implies we have either $t-3=0$ or $t+1=0$

The first equation can be solved by adding 3 on both sides: $t=3$ .

The second equation can be solved by subtracting 1 on both sides: $t=-1$ .

So when $t=3$ seconds, is when the rocket has hit the ground.

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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 $t_{\alpha/2}= 2.79$

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