All categories

4 years ago

Which statement is true regarding the functions of the graph?

Mathematics

2 answers:

Oksana_A [137]4 years ago

4 0

I think it’s f(3)= g(6)

pantera1 [17]4 years ago

4 0

Answer:

Its B

Step-by-step explanation:

I took the test

You might be interested in

The school band has a goal of raising $800 for some new uniforms. If they have raised $512 so far, what percent of their goal ha

uranmaximum [27]

64% divide 512 by 800

3 0

3 years ago

Read 2 more answers

5. Order these from least to greatest. 1/2,4/9, 2/3,5/6

stiks02 [169]

Answer: this is the correct order: 4/9,1/2,2/3,5/6

Step-by-step explanation:

7 0

3 years ago

In a study to compare two different corrosion inhibitors, specimens of stainless steel were immersed for four hours in a solutio

djverab [1.8K]

Answer:

$(242-220) -1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 10.862$

$(242-220) +1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 33.138$

And the 95% confidence interval for the difference in the means is given by: $10.862 \leq \mu_A -\mu_B \leq 33.138$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_A = 242$ sample mean for inhibitor A

$s_A = 20$ sample standard deviation for inhibitor A

$n_A = 47$ sample size for A

$\bar X_B = 220$ sample mean for inhibitor B

$s_B = 31$ sample standard deviation for inhibitor B

$n_B = 42$ sample size for A

Solution to the problem

For this case the confidence interval for the difference of means is given by:

$(\bar X_A -\bar X_B) \pm t_{\alpha/2} \sqrt{\frac{s^2_A}{n_A} +\frac{s^2_B}{n_B}}$

The degrees of freedom are given by:

$df = n_A +n_B -2 = 47+42-2= 87$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,87)".And we see that $t_{\alpha/2}=1.988$

And replacing we got:

$(242-220) -1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 10.862$

$(242-220) +1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 33.138$

And the 95% confidence interval for the difference in the means is given by: $10.862 \leq \mu_A -\mu_B \leq 33.138$

4 0

3 years ago

Find the midpoint M of the line segment joining the points S = (-3, -6) and T = (5, 4). M =

grandymaker [24]

Midpoint is simply the half of the distance between two points. In order to find this we will need to find the distance first and then find the midpoint... Let's start solving!~

$\red{ \rule{35pt}{2pt}} \orange{ \rule{35pt}{2pt}} \color{yellow}{ \rule{35pt} {2pt}} \green{ \rule{35pt} {2pt}} \blue{ \rule{35pt} {2pt}} \purple{ \rule{35pt} {2pt}}$