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butalik [34]
3 years ago
10

PLEASE HELP WITH PROBLEM

Mathematics
1 answer:
Crank3 years ago
4 0
P=12,500×(1+0.23)^(4)
P=28,610.83
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A soma de três números naturais e consecutivos é 636. Essa soma é:
agasfer [191]

Answer:

El anwer es A porque es el consecutivos, tambien abla en los the espanol

Step-by-step explanation:

8 0
3 years ago
Let → a = ⟨ 1 , − 3 ⟩ and → b = ⟨ − 3 , k ⟩. Find k so that →a and → b will be orthogonal (form a 90 degree angle).
Mice21 [21]
If a b is orthogonal , a•b will be 0
a•b=1x-3+-3k=0
so k=-1
7 0
3 years ago
What is square root of 126 + square root of 56 in standard form
Feliz [49]
\sqrt{126}= 3\sqrt{14} \\ \\ \sqrt{56}=2\sqrt{14} \\ \\ \\ \sqrt{126}+ \sqrt{56}= 3\sqrt{14}+2\sqrt{14}= 5\sqrt{14}
6 0
3 years ago
Read 2 more answers
Use the drawing tool(s) to form the correct answer on the provided graph.
NikAS [45]

Answer:

Here's one way to do it

Step-by-step explanation:

1. Solve the inequality for y

5x - y > -3

-y > -5x - 3

y < 5x + 3

2. Plot a few points for the "y =" line

I chose

\begin{gathered}\begin{array}{rr}\mathbf{x} & \mathbf{y} \\-2 & -7 \\-1 & -2 \\0 & 3 \\1 & 8 \\2 & 13 \\\end{array}\end{gathered}

x

−2

−1

0

1

2

y

−7

−2

3

8

13

You should get a graph like Fig 1.

3. Draw a straight line through the points

Make it a dashed line because the inequality is "<", to show that points on the line do not satisfy the inequality.

See Fig. 2.

4. Test a point to see if it satisfies the inequality

I like to use the origin,(0,0), for easy calculating.

y < 5x + 3

0 < 0 + 3

0 < 3. TRUE.

The condition is TRUE.

Shade the side of the line that contains the point (the bottom side).

And you're done (See Fig. 3).

6 0
3 years ago
The Quality Control Department employs five technicians during the day shift. Listed below is the number of times each technicia
sveta [45]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data :

Technician __Shutdown

Taylor, T___4

Rousche, R _ 3

Hurley, H__ 3

Huang, Hu___2

Gupta, ___ 5

The Numbe of samples of 2 possible from the 5 technicians :

We use combination :

nCr = n! ÷ (n-r)!r!

5C2 = 5!(3!)2!

5C2 = (5*4)/2 = 10

POSSIBLE COMBINATIONS :

TR, TH, THu, TG, RH, RHu, RG , HHu, HG, HuG

Sample means :

TR = (4+3)/2 = 3.5

TH = (4+3)/2 = 3.5

THu = (4+2) = 6/2 = 3

TG = (4 + 5) = 9/2 = 4.5

RH = (3+3) = 6/2 = 3

RHu = (3+2) /2 = 2.5

RG = (3 + 5) = 8/2 = 4

HHu = (3+2) = 2.5

HG = (3+5) = 8/2 = 4

HuG = (2+5) / 2 = 3.5

Mean of sample mean (3.5+3.5+3+4.5+3+2.5+4+2.5+4+3.5) / 10 = 3.4

Population mean :

(4 + 3 + 3 + 2 + 5) / 5 = 17 /5 = 3.4

Population Mean and mean of sample means are the same.

This distribution should be approximately normal.

7 0
3 years ago
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