2 years ago

Draw two possible rectangles for each perimeter. Find the area 20 units

Mathematics

1 answer:

Gre4nikov [31]2 years ago

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Draw a triangle with opposite side 5 and 4 and one with opposite sides 10 and 2

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[(x^5)]^7]^4<br><br> simplify the expression

lina2011 [118]

$[(x^{5}) ^7]^4 = x^{5*7*4} = x^{140}$

ok done. Thank to me :>

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1 year ago

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In-s [12.5K]

Answer: Sorry can’t help

Step-by-step explanation:

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2 years ago

Find the side length of a cube when the volume is 216 ft cubed. A) 72 ft B) 6 ft C) 108 ft D) 219 ft

andrew11 [14]

If its 216 cubic inches it would be A)

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2 years ago

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Right ∆ABC has coordinates A(-7, 3), B( -7, 10), and C(-1, 3). The triangle is reflected over the x - axis and then reflected ag

Mrrafil [7]

Vertex A’ would be (7, -3)

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2 years ago

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A survey found that women's heights are normally distributed with mean 63.2 in. and standard deviation 2.4 in. The survey also

Effectus [21]

Answer: 99.51%

Step-by-step explanation:

Given : A survey found that women's heights are normally distributed.

Population mean : $\mu =63.2 \text{ inches}$

Standard deviation: $\sigma= 2.4\text{ inches}$

Minimum height = 4ft. 9 in.= $4\times12+9\text{ in.}=57\text{ in.}$

Maximum height = 6ft. 2 in.= $6\times12+2\text{ in.}=74\text{ in.}$

Let x be the random variable that represent the women's height.

z-score : $z=\dfrac{x-\mu}{\sigma}$

For x=57, we have

$z=\dfrac{57-63.2}{2.4}\approx-2.58$

For x=74, we have

$z=\dfrac{74-63.2}{2.4}\approx4.5$

Now, by using the standard normal distribution table, we have

The probability of women meeting the height requirement :-

$P(-2.58$

Hence, the percentage of women meeting the height requirement = 99.51%

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2 years ago