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Vitek1552 [10]
3 years ago
5

360 increased by 20% decreased by 50

Mathematics
1 answer:
castortr0y [4]3 years ago
8 0
370

20%= 0.2

0.2(360)=72

360+72=432

432-50=382
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Find the area of the triangle defined by the coordinates (7, 1), (0, 10), and (9, 4). (To the nearest tenth)
slega [8]

Answer:

  C)  19.5 square units

Step-by-step explanation:

The area from coordinates can be computed as ...

  A = (1/2)|x1(y2 -y3) +x2(y3 -y1) +x3(y1 -y2)|

  = (1/2)|7(10 -4) +0(4 -1) +9(1-10)| = (1/2)|7·6 +9·(-9)|

  = (1/2)|-39| = 39/2 = 19.5

The area of the triangle is 19.5 square units.

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3 years ago
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K.Brew sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet. R
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Answer:

Step-by-step explanation:

The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

Where

x1 = mean sale amount for mail order sales = 82.70

x2 = mean sale amount for internet sales = 66.9

s1 = sample standard deviation for mail order sales = 16.25

s2 = sample standard deviation for internet sales = 20.25

n1 = number of mail order sales = 17

n2 = number of internet sales = 10

For a 99% confidence interval, we would determine the z score from the t distribution table because the number of samples are small

Degree of freedom =

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z = 2.787

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The doubling period of a bacterial population is 15 minutes. At time t= 80 minutes, the bacterial population was 90000​
Papessa [141]

Answer:

here i finished!

hope it helps yw!

Step-by-step explanation:

The doubling period of a bacterial population is 15 minutes.

At time t = 90 minutes, the bacterial population was 50000.

Round your answers to at least 1 decimal place.

:

We can use the formula:

A = Ao*2^(t/d); where:

A = amt after t time

Ao = initial amt (t=0)

t = time period in question

d = doubling time of substance

In our problem

d = 15 min

t = 90 min

A = 50000

What was the initial population at time t = 0

Ao * 2^(90/15) = 50000

Ao * 2^6 = 50000

We know 2^6 = 64

64(Ao) = 50000

Ao = 50000/64

Ao = 781.25 is the initial population

:

Find the size of the bacterial population after 4 hours

Change 4 hr to 240 min

A = 781.25 * 2^(240/15

A = 781.25 * 2^16

A= 781.25 * 65536

A = 51,199,218.75 after 4 hrs

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How can I solve each system of liner equations by substitution?
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I hope you can read cursive and I hope this helps! (:

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