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3 years ago

will a 13 inch rusty metal bar fit into a box (rectangular prism) with dimensions 9 inches by 8 inches by 3 inches?

Mathematics

2 answers:

den301095 [7]3 years ago

4 0

Answer:

Answer:

Bar will not fit in the box.

Step-by-step explanation:

Given:

Length = 9 in

Width = 8 in

Height = 3 in

Length of bar = 13 in

Find:

Bar fit in box or not

Computation:

If length of bar is equal to length of diagonal than it will fit in box.

Diagonal of a cuboid = √l² + b² + h²

Diagonal of a cuboid = √9² + 8² + 3²

Diagonal of a cuboid = √81 + 64 + 9

Diagonal of a cuboid = √154

Diagonal of a cuboid = 12.40 in

Diagonal of a cuboid < Length of bar

So,

Bar will not fit in the box.

Step-by-step explanation:

wariber [46]3 years ago

3 0

Answer:

Bar will not fit in the box.

Step-by-step explanation:

Given:

Length = 9 in

Width = 8 in

Height = 3 in

Length of bar = 13 in

Find:

Bar fit in box or not

Computation:

If length of bar is equal to length of diagonal than it will fit in box.

Diagonal of a cuboid = √l² + b² + h²

Diagonal of a cuboid = √9² + 8² + 3²

Diagonal of a cuboid = √81 + 64 + 9

Diagonal of a cuboid = √154

Diagonal of a cuboid = 12.40 in

Diagonal of a cuboid < Length of bar

So,

Bar will not fit in the box.

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4 years ago

Cidnee is weighing combinations of geometric solids. She found that 4 cylinders and 5 prisms weigh 32 ounces and that 1 cylinder

vitfil [10]

Answer:

The weights of each cylinder and prism are 3 and 4 ounces, respectively.

Step-by-step explanation:

Let be $x$ and $y$ the masses of a cylinder and a prism, measured in ounces, respectively. After a careful reading of the statement we get the following linear equations by interpretation:

i) She found that 4 cylinders and 5 prisms weigh 32 ounces:

$4\cdot x +5\cdot y = 32\,oz$ (Eq. 1)

ii) And that 1 cylinder and 8 prisms weigh 35 ounces:

$x + 8\cdot y = 35\,oz$ (Eq. 2)

Now we solve the system of linear equations algebraically:

From (Eq. 2):

$x = 35 - 8\cdot y$

(Eq. 2) is (Eq. 1):

$4\cdot (35-8\cdot y) + 5\cdot y = 32$

$140-32\cdot y +5\cdot y = 32$

$32\cdot y - 5\cdot y = 140-32$

$27\cdot y = 108$

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3 years ago

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QveST [7]

Answer:

We conclude that the mean nicotine content is less than 31.7 milligrams for this brand of cigarette.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 31.7 milligrams

Sample mean, $\bar{x}$ = 28.5 milligrams

Sample size, n = 9

Alpha, α = 0.05

Sample standard deviation, s = 2.8 milligrams

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 31.7\text{ milligrams}\\H_A: \mu < 31.7\text{ milligrams}$

We use One-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{28.5 - 31.7}{\frac{2.8}{\sqrt{9}} } = -3.429$

Now, $t_{critical} \text{ at 0.05 level of significance, 8 degree of freedom } = -1.860$

Since,

$t_{stat} < t_{critical}$

We fail to accept the null hypothesis and accept the alternate hypothesis. We conclude that the mean nicotine content is less than 31.7 milligrams for this brand of cigarette.

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3 years ago