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Usimov [2.4K]
2 years ago
13

Please answer thisssss

Mathematics
1 answer:
crimeas [40]2 years ago
8 0

Answer:

A, C and E

Step-by-step explanation:

In the graph, it looks like there are no outliers, as the points are all generally in one line, it looks linear and it has a negative association because as the x value increases, the y value decreases.

Hope this helps!

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What is 3(5x^2-9x) evaluated for x=-3
Studentka2010 [4]

Answer:

216

Step-by-step explanation:

put -3 where x is

3 ( 5 -3^2 - 9(-3) )

3 ( 45 + 27 )

3 (72)

216

4 0
2 years ago
Read 2 more answers
J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.
melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

                      P.Q.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~ t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean for mail-order sales = $74.50

\bar X_2 = sample mean for internet sales = $84.40

s_1 = sample standard deviation for mail-order purchases = $17.25

s_2 = sample standard deviation for internet purchases = $21.25

n_1 = sample of sales receipts for mail-order purchases = 16

n_2 = sample of sales receipts for internet purchases = 9

Also,  s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }  =  \sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} } = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by (\mu_1-\mu_2).

Now, 99% confidence interval for (\mu_1-\mu_2) is given by;

             = (\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_)  \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

          = (74.50-84.40) \pm (2.807  \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})

          = [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0
3 years ago
How many feet would you cover if you walked for 30 minutes?​
serg [7]

Answer: 2.0 miles= 10560 feet

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Segment BD is an altitude of triangle ABC. Find the area of the triangle. Triangle ABC with altitude BD is shown. Point A is at
pogonyaev
Where's the "triangle with alt. BD?"  This problem can be solved without the diagram, but the solution would be easier with it.

BD is the altitude.  Find the length of BD by finding the dist. between (-1,4) and (2,4); it is 2-(-1), or 3.  |BD| = 3.

I've graphed the triangle myself and have found that the "base" of the triangle is the vertical line thru (2,1) and (2,6); its length is 6-1, or 5.

Thus, the area of this triangle is  A = (b)(h) / 2, or  A = (5)(3) / 2 = 10/3 square inches.
7 0
3 years ago
Read 2 more answers
Anybody know the answer ?
IgorC [24]
Im pretty sure its -125/51
6 0
3 years ago
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