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natima [27]
3 years ago
12

Pls help meeeeeeee with this

Mathematics
1 answer:
hammer [34]3 years ago
4 0
:):(:(:):)(:(:$:):(:(:(:):)3():):);;
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Please help me <br>why - 9/5-1=-14/5​
Jobisdone [24]

Answer:

Step-by-step explanation:

Because you have to take the L.C.M ,

-9/5 - 1

=( -9-5)/5

= -14/5

please  

Brain-list it or support me at my U-Tube channel " ZK SOFT&GAMING " I will be thankful

6 0
3 years ago
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I'll go on a date with anyone who can answer this and help me...?
vesna_86 [32]

Answer:

Chill is this a multiple choice?

7 0
3 years ago
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
Which set of numbers could represent the lengths of the sides of a right triangle? 16, 32, 36 5, 12, 13 6, 7, 8 8, 12, 16
Elden [556K]

5,12,13

5X5=25

12X12=144

25+144=169

answer: square root 169=13

6 0
3 years ago
Line $\ell_1$ is horizontal and passes through $(1,1)$. Line $\ell_2$ is vertical and passes through $(2,2)$. Find the point of
borishaifa [10]

Answer:

(2,1)

Step by step explanation:      

Please find the attachment.

We have been given that line l_1 is horizontal and passes through point (1,1). Line l_2 is vertical and passes through point (2,2). We are asked to find point of intersection of line l_1 and line l_2.  

Line l_1 is a horizontal line so y-coordinate will be 1 for any value of x. Line l_2 is a vertical line, so x-coordinate will be 2 for any value of y.

Therefore, intersection point of both lines will include x-coordinate of line l_2 and y-coordinate of line l_1 that is (2,1).

5 0
3 years ago
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