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liraira [26]
2 years ago
11

A regular-size box of cereal measures 3 1/2 inches by 8 1/2 inches by 15 inches. The manufacturer also sells an individual-size

box that has a volume that is 1 1/0 of the volume of the regular-size box.
What is the volume of the individual-size box of cereal?

Enter your answer as a mixed number in simplest form by filling in the boxes.
Mathematics
1 answer:
Phoenix [80]2 years ago
4 0

The individual-size box of cereal is a rectangular prism with a volume of 44.625 inches³.

<h3>What is the volume of a rectangular prism?</h3>

The volume of rectangular prism of length l, width w and height h is given by:

V = lwh.

In this problem, the standard dimensions are:

  • l = 3(1/2) = 3 + 0.5 = 3.5 inches.
  • w = 8(1/2) = 8 + 0.5 = 8.5 inches.
  • h = 15 inches.

The individual-size box has a 1/10 of the volume of the original box, hence:

V = 0.1 x 3.5 x 8.5 x 15 = 44.625 inches³.

More can be learned about the volume of a rectangular prism at brainly.com/question/17223528

#SPJ1

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Answer:Average error will be less

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3 years ago
PLEASE HELP!!<br><br> which graph represents the function f(x) = -|x| - 2
Nonamiya [84]

Answer:

The graph on the far right

Step-by-step explanation:

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2 years ago
What is the equation of M=3,x=4,and b=9
natita [175]

Answer:

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Step-by-step explanation:

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Please be more specific so that you get the answer you need

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4 years ago
Suppose you are investigating the relationship between two variables, traffic flow and expected lead content, where traffic flow
prisoha [69]

Answer:

The answers is " Option B".

Step-by-step explanation:

CI=\hat{Y}\pm t_{Critical}\times S_{e}

Where,

\hat{Y}= predicted value of lead content when traffic flow is 15.

\to df=n-1=8-1=7

 95\% \ CI\  is\  (463.5, 596.3) \\\\\hat{Y}=\frac{(463.5+596.3)}{2}\\\\

     =\frac{1059.8}{2}\\\\=529.9

Calculating thet-critical valuet_{ \{\frac{\alpha}{2},\ df \}}=-2.365

The lower predicted value =529.9-2.365(Se)

463.5=529.9-2.365(Se)\\\\529.9-463.5=2.365(Se)\\\\66.4=2.365(Se)\\\\Se=\frac{66.4}{2.365} \\\\Se=28.076

When 99\% of CI use as the expected lead content: \to 529.9\pm t_{0.005,7}\times 28.076 \\\\=(529.9 \pm 3.499 \times 28.076)\\\\=(529.9 \pm 98.238)\\\\=(529.9-98.238, 529.9+98.238)\\\\=(431.662, 628.138)\\\\=(431.6, 628.1)

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