Please help me 1,2,3 or 4 need ASAP.

What is the equation has intercepts at x(1, 0, 0), y(0, -1,0), and z(0, 0, 2)

bulgar [2K]

In intercept form, the plane that has these intercepts is ...

... x/(x-intercept) + y/(y-intercept) + z/(z-intercept) = 1

... x/1 + y/(-1) + z/2 = 1

... 2x -2y +z = 2 . . . . . in standard form

3 0

3 years ago

Consider the following hypothesis test:

Gnom [1K]

Answer:

a. z=3.09. Yes, it can be concluded that the population mean is greater than 50.

b. z=1.24. No, it can not be concluded that the population mean is greater than 50.

c. z=2.22. Yes, it can be concluded that the population mean is greater than 50.

Step-by-step explanation:

We have a hypothesis test for the mean, with the hypothesis:

$H_0: \mu\leq50\\\\H_a:\mu> 50$

The sample size is n=55 and the population standard deviation is 6.

The significance level is 0.05.

We can calculate the standard error as:

$\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{55}}=0.809$

For a significance level of 0.05, the critical value for z is zc=1.644. If the test statistic is bigger than 1.644, the null hypothesis is rejected.

a. If the sample mean is M=52.5, the test statistic is:

$z=\dfrac{M-\mu}{\sigma_M}=\dfrac{52.5-50}{0.809}=\dfrac{2.5}{0.809}=3.09$

The null hypothesis is rejected, as z>zc and falls in the rejection region.

b. If the sample mean is M=51, the test statistic is:

$z=\dfrac{M-\mu}{\sigma_M}=\dfrac{51-50}{0.809}=\dfrac{1}{0.809}=1.24$

The null hypothesis failed to be rejected, as z<zc and falls in the acceptance region.

c. If the sample mean is M=51.8, the test statistic is:

$z=\dfrac{M-\mu}{\sigma_M}=\dfrac{51.8-50}{0.809}=\dfrac{1.8}{0.809}=2.22$

The null hypothesis is rejected, as z>zc and falls in the rejection region.

8 0

3 years ago

Best has knitted 5/8 of the scarf she is making has she needed more than one half of the scarf can't find the equivalent fractio

BabaBlast [244]

To compare 5/8 and 1/2, you will need to create fractions that have the same denominator. 1/2 is the same as 4/8.

5/8 is greater than 4/8, so she had knitted more than 1/2.

4 0

3 years ago

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question. Listed below

mafiozo [28]

Answer:

a) $\bar X = 369.62$

b) $Median=175$

c) $Mode =450$

With a frequency of 4

d) $MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

<u>e)</u> $s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

Step-by-step explanation:

We have the following data set given:

49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000

Part a

The mean can be calculated with this formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

Replacing we got:

$\bar X = 369.62$

Part b

Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

$Median=175$

Part c

The mode is the most repeated value in the sample and for this case is:

$Mode =450$

With a frequency of 4

Part d

The midrange for this case is defined as:

$MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

Part e

For this case we can calculate the deviation given by:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

5 0

3 years ago

00:00

Daniel [21]

Answer:

13 packages of Granola

Step-by-step explanation:

74.25 - 6 = 68.25

68.25 divided by 5.25= 13

3 0

3 years ago