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eimsori [14]
1 year ago
11

Points A, B, and C, form a triangle. The distance between point A and point B is 15 yards. The distance between point B and poin

t C is 25 yards. Pete walks directly from point A to point C, without passing through point B. What is the direct distance from A to C?
How far would Pete walk if he went from A to B to C?
yards
The direct distance from A to C is more than
yards.
The inequality w <
represents the distance, w, that Pete might save by taking the direct path.
Mathematics
2 answers:
DIA [1.3K]1 year ago
8 2

Answer:

(a)

A to B to C =40 yards

(b)

w<40

Step-by-step explanation:

We are given

The distance between point A and point B is 15 yards

AB=15 yards

The distance between point B and point C is 25 yards

BC=25 yards

(a)

A to B to C = AB+BC

A to B to C =15+25

A to B to C =40 yards

(b)

direct distance between A to C  or AC

It must be less  than distance between A to B to C

AC< AB+BC

AC< 15+25

AC< 40 yards

w<40


Guest
1 year ago
tot
Guest
1 year ago
totally wrong but thanks for trying
ozzi1 year ago
9 0
The answers are 40, 10, 30
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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.

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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].

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matrenka [14]

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where n is the number of sides and since there are  sides:

total interior angle = 360°

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