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3 years ago

7

Derek is playing a prank on Grace and hid her pencil inside a tissue box that has a width of 4 inches, a length of 8 inches, and

a height of 4 inches. The pencil fits perfectly from the upper left corner to the bottom right corner of the box. What is the length of Grace’s pencil to the nearest tenth of an inch?

2 answers:

ohaa [14]3 years ago

8 0

Answer:

9.8 Inch

Step-by-step explanation:

The distance from the upper left corner to the bottom right corner of the box is the longest diagonal of the box.

For a rectangular prism:

Length of the Longest diagonal $=\sqrt{ l^2+h^2+w^2}$

Therefore:

Length of the pencil $=\sqrt{ 8^2+4^2+4^2}$

$=\sqrt{96}\\=9.8$ inch(correct to the nearest tenth of an inch)$

m_a_m_a [10]3 years ago

4 0

Answer: length of pencil is 9.8 inches.

Step-by-step explanation:

The pencil fits perfectly from the upper left corner to the bottom right corner of the box. This is the diagonal of the box(not the diagonal of a face). To determine the length of the pencil, we would determine the length of the diagonal of the box. The formula formula is

d = √(l² + w² + h²)

Where

d = diagonal = length of pencil

w = width = 4 inches

h = height = 4 inches

l = length = 8 inches

Therefore,

d = √(4² + 4² + 8²) = √96

d = 9.8 inches

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2(a³+b²+11)+1(3+a³+b+11)

Ann [662]

Answer:

$2(a^3+b^2+11)+1(3+a^3+b+11)=3a^3+2b^2+b+36$

Step-by-step explanation:

I assume that you need simplification of the given expression.

The given expression is:

$2(a^3+b^2+11)+1(3+a^3+b+11)$

Using distributive property and multiplying 2 inside the parenthesis and 1 inside the other parenthesis. This gives,

$2(a^3+b^2+11)=2\times a^3+2\times b^2+2\times 11\\2(a^3+b^2+11)=2a^3+2b^2+22\\\\1(3+a^3+b+11)=1\times 3+1\times a^3+1\times b+1\times 11\\1(3+a^3+b+11)=3+a^3+b+11=a^3+b+14$

Therefore, $2(a^3+b^2+11)+1(3+a^3+b+11)$ is equal to:

$2a^3+2b^2+22+a^3+b+14$

Now, combining like terms using the commutative property of addition, we get:

$=(2a^3+a^3)+2b^2+b+(22+14)\\=3a^3+2b^2+b+36$

Therefore, the simplified form is $3a^3+2b^2+b+36$

8 0

3 years ago

If the average yield of cucumber acre is 800 kg, with a variance 1600 kg, and that the amount of the cucumber follows the normal

lorasvet [3.4K]

Answer:

a

The percentage is

$P(x_1 < X < x_2 ) = 51.1 \%$

b

The probability is $P(Z > 2.5 ) = 0.0062097$

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 800$

The variance is $var(x) = 1600 \ kg$

The range consider is $x_1 = 778 \ kg \ x_2 = 834 \ kg$

The value consider in second question is $x = 900 \ kg$

Generally the standard deviation is mathematically represented as

$\sigma = \sqrt{var (x)}$

substituting value

$\sigma = \sqrt{1600}$

$\sigma = 40$

The percentage of a cucumber give the crop amount between 778 and 834 kg is mathematically represented as

$P(x_1 < X < x_2 ) = P( \frac{x_1 - \mu }{\sigma} < \frac{X - \mu }{ \sigma } < \frac{x_2 - \mu }{\sigma } )$

Generally $\frac{X - \mu }{ \sigma } = Z (standardized \ value \ of \ X)$

So

$P(x_1 < X < x_2 ) = P( \frac{778 - 800 }{40} < Z< \frac{834 - 800 }{40 } )$

$P(x_1 < X < x_2 ) = P(z_2 < 0.85) - P(z_1 < -0.55)$

From the z-table the value for $P(z_1 < 0.85) = 0.80234$

and $P(z_1 < -0.55) = 0.29116$

So

$P(x_1 < X < x_2 ) = 0.80234 - 0.29116$

$P(x_1 < X < x_2 ) = 0.51118$

The percentage is

$P(x_1 < X < x_2 ) = 51.1 \%$

The probability of cucumber give the crop exceed 900 kg is mathematically represented as

$P(X > x ) = P(\frac{X - \mu }{\sigma } > \frac{x - \mu }{\sigma } )$

substituting values

$P(X > x ) = P( \frac{X - \mu }{\sigma } >\frac{900 - 800 }{40 } )$

$P(X > x ) = P(Z >2.5 )$

From the z-table the value for $P(Z > 2.5 ) = 0.0062097$

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