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hjlf
2 years ago
8

The area for this shape.

Mathematics
2 answers:
Bogdan [553]2 years ago
8 0

Answer:

Step-by-step explanation:

kirill [66]2 years ago
3 0
Answer:

240 cm

Step-by-step explanation:

10 • 18 = 180

18 - 6 = 12

12 • 5 = 60/2 = 30

12 • 5 = 60/2 = 30

180 + 30 + 30 = 240
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The balance on a car loan after 4 years is $8,996.32. The interest rate is 5.6% compounding annually. What was the initial value
Slav-nsk [51]
You would get 2015.17568 as the interest rate i believe 
6 0
3 years ago
Translate the following into verbal phrases. 3x-4
nadezda [96]
Answer:
Three x negative four
7 0
3 years ago
Try It!
Wittaler [7]

Answer:

91.46m

Step-by-step explanation:

The length of 100 yard football field in meters.

From the dimension given,

1 yard = 3 feet

Therefore,

100 yard = x feet

1 yard = 3 feet

100 yard = x feet

X = (100 × 3) / 1

X = 300 feet.

But,

1 metre = 3.28feet

X metre = 300 feet

X = (300 × 1) / 3.28

X = 91.46m

Therefore, 100 yard football field is equal to 91.46m

6 0
2 years ago
Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the corre
fenix001 [56]

Answer:

x^6y^3

Step-by-step explanation:

<u>Given polynomial:</u>

  • 8x^6y^5 - 3x^8y^3

<u>Get prime factors:</u>

  • 2*2*2*x*x*x*x*x*x*y*y*y*y*y - 3*x*x*x*x*x*x*x*x*y*y*y

<u>Common factors of the two terms:</u>

  • x*x*x*x*x*x*y*y*y = x^6y^3

<u>So</u>

  • x^6y^3(8y^2 - 3x^2)

x^6y^3 is the GCF of this polynomial

4 0
3 years ago
A population of monkeys' tail lengths is normally distributed with a mean of 25 cm with a standard deviation of 8 cm. I am prepa
Crank

Answer:

<em>The probability that the mean of my sample will be between 24 and 25 cm</em>

<em>P(24 ≤X⁻≤25) = 0.4772</em>

Step-by-step explanation:

<u><em>Step(i)</em></u>:-

<em>Given mean of the Population  'μ'= 25c.m</em>

<em>Given standard deviation of the Population 'σ' = 8c.m</em>

<em>Given sample size 'n' = 256</em>

<em>Let X₁ = 24</em>

<em></em>Z_{1}  = \frac{x_{1}-mean }{\frac{S.D}{\sqrt{n} } } = \frac{24-25}{\frac{8}{\sqrt{256} } } =  -2<em></em>

<em>Let X₂ = 25</em>

<em></em>Z_{2}  = \frac{x_{2}-mean }{\frac{S.D}{\sqrt{n} } } = \frac{25-25}{\frac{8}{\sqrt{256} } } = 0<em></em>

<u><em>Step(ii)</em></u><em>:-</em>

<em>The probability that the mean of my sample will be between 24 and 25 cm</em>

<em>P(24 ≤X⁻≤25) = P(-2≤ Z ≤0)</em>

                     = P( Z≤0) - P(Z≤-2)

                     = 0.5 + A(0) - (0.5- A(-2))

                     = A(0) + A(2)        ( ∵A(-2) =A(2)

                     = 0.000+ 0.4772

                     = 0.4772

<u><em>Final answer</em></u>:-

<em>The probability that the mean of my sample will be between 24 and 25 cm</em>

<em>P(24 ≤X⁻≤25) = 0.4772</em>

                     

4 0
3 years ago
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