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andrew11 [14]
2 years ago
14

The ratio of the lengths of the sides of △ABC is 3:6:7. M, N, and K are the midpoints of the sides. Perimeter of △MNK equals 7.4

in. Find the lengths of the sides of △ABC.
Mathematics
1 answer:
artcher [175]2 years ago
8 0

Answer:

AB=2.775

BC=5.55

CA=6.475

Step-by-step explanation:

Since midpoints split their sides in half, we can see that the triangle MNK formed by the midpoints will be half the perimeter of the triangle ABC. Since P of MNK = 7.4, we know that the perimeter of ABC = 7.4 * 2, which is 14.8. Now we can split the 14.8 so that it follows the ratio.

3+6+7=16

14.8/16=0.925

AB=0.925*3=2.775

BC=0.925*6=5.55

CA=0.925*7=6.475

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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.

          = (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
2 years ago
Jordan is making gifts for volunteers and orders 4,580 personalized M&Ms. She puts 34 M&Ms in each gift. How many gifts
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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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1 year ago
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2 years ago
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Answer:

Step-by-step explanation:

The given quadratic equation is

2x^2+3x-8 = 0

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x = -b ± √b^2 - 4ac]/2a

from the equation,

a = 2

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It becomes

x = [- 3 ± √3^2 - 4(2 × -8)]/2×2

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x = (-3 + 8.544) /4 or x = (-3 - 8.544) / 4

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x = 1.386 or x = - 2.886

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