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3 years ago

Whoever is good at math, please help me!

Mathematics

1 answer:

Kruka [31]3 years ago

4 0

Answer:

I think it is 62.4 km.

Step-by-step explanation:

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Kierston needs to paint a wall of the youth center. She knows that 1 can of paint covers an area of 2.2 square meters. Kierston

Step2247 [10]

Answer:

See Explanation

Step-by-step explanation:

Given

$1\ can = 2.2m^2$

Required

Determine the number of cans for the wall

The dimension of the wall is not given. So, I will use the following assumed values:

$Length=20m$

$Width = 44m$

First, calculate the area of the wall

$Area = Length * Width$

$Area = 20m * 44m$

$Area = 880m^2$

If $1\ can = 2.2m^2$

Then $x = 880m^2$

Cross Multiply:

$x * 2.2m^2 = 1 * 880m^2$

$x * 2.2m^2 = 880m^2$

$x * 2.2 = 880$

Make x the subject

$x = \frac{880}{2.2}$

$x = 400$

400 cans using the assume dimensions.

So, all you need to to is, get the original values and follow the same steps

4 0

2 years ago

Determine which postulate can be used

Kruka [31]

Answer:

SAS postulate

Step-by-step explanation:

The triangles have two congruent sides and one congruent angle. The congruent angle is the included angle. This meets SAS criteria.

<u>Hope this helps :-)</u>

3 0

2 years ago

Help me find the length of BC?

sergiy2304 [10]

Is This For Geometry?

6 0

2 years ago

Find the width w of a box with volume V = 120 cm3, l = 3 cm and h = 5 cm.

LenKa [72]

Answer:

Step-by-step explanation:

the volume of a box is L x W x H

so if you take the length x height, you can divide the volume by that to get your width.

hope this helps!! :)

8 0

2 years ago

Read 2 more answers

Suppose the weights of farmer carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1

denis23 [38]

Given that <span>the weights of farmer carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1 ounces.

The probability of a normally distributed data between two values (a, b) is given by:

$P(a\ \textless \ X\ \textless \ b)=P\left(z\ \textless \ \frac{b-\mu}{\sigma/\sqrt{n}} \right)-P\left(z\ \textless \ \frac{a-\mu}{\sigma/\sqrt{n}} \right) \\ \\ =P(7.5\ \textless \ X\ \textless \ 8.5)=P\left(z\ \textless \ \frac{8.5-8.0}{1.1/\sqrt{6}} \right)-P\left(z\ \textless \ \frac{7.5-8.0}{1.1/\sqrt{6}} \right) \\ \\ =P(z\ \textless \ 1.113)-P(-1.113)=0.8672-0.1328=0.7344$ </span>

6 0

3 years ago