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bazaltina [42]
4 years ago
15

A trapezoidal mural with bases 11m and 9m and height 6m needs to be painted. What is the cost of painting it if a box of paint i

n any color is $16 and with each box can be painted area of 12m2.
Mathematics
2 answers:
mel-nik [20]4 years ago
8 0

Answer:

80 doolers

Step-by-step explanation:

Dovator [93]4 years ago
4 0
Answer: The total cost to paint the mural will be $80.

First, we need to find the area of the trapezoid. The formula is one-half times the height times the sum of the bases.

0.5 x 6 (9 + 11) = 60 m^2

Each can will cover 12 m^2, so divide 60 by 12 to find that we need 5 cans.

The cost of each can is $16, so 5 x 16 = $60.
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What is 2/3 divided by 5?<br> Show your work
Zolol [24]

Answer:

2/15

Step-by-step explanation:

2/3 / 5/1  =  

2/3 x 1/5 = 2/15

4 0
3 years ago
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REALLY NEED HELP
artcher [175]

Answer:

Y=-5/3x

Step-by-step explanation:

Use the slope formula and slope-intercept form y=mx+b to find the equation.

3 0
3 years ago
The lengths of pregnancies in a small rural village are normally distributed with a mean of 265 days and a standard deviation of
MakcuM [25]

Answer:

the middle 50% of most lengths of pregnancies ranges between 255.62 days and 274.38 days

Step-by-step explanation:

Given that :

Mean  = 265

standard deviation = 14

The formula for calculating the z score is z = \dfrac{x -\mu}{\sigma}

x = μ + σz

At middle of 50% i.e 0.50

The critical value for z_{\alpha/2} = z_{0.50/2}

From standard normal table

z_{0.25}= + 0.67  or -0.67  

So; when z = -0.67

x = μ + σz

x = 265 + 14(-0.67)

x = 265 -9.38

x = 255.62

when z = +0.67

x = μ + σz

x = 265 + 14 (0.67)

x = 265 + 9.38

x = 274.38

the middle 50% of most lengths of pregnancies ranges between 255.62 days and 274.38 days

3 0
3 years ago
I ONLY NEED HELP ON NUMBER 2
REY [17]
So, the area of a circle is a=pi r^2
so do that and get a=pi(4)
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4 0
3 years ago
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6.75 + StartFraction 3 Over 8 EndFraction x = 13 and one-fourth
amid [387]
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3 years ago
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