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Mazyrski [523]
3 years ago
12

Does 6% of a pound weigh more than an ounce

Mathematics
2 answers:
julia-pushkina [17]3 years ago
8 0

Answer:

6% of a pound does not weigh more than 1 ounce.

Step-by-step explanation:

First, we will calculate 6% (0.06) of 1 pound.

0.06 × 1 lb = 0.06 lb

1 pound is equal to 16 ounces. 0.06 pounds, expressed in ounces is:

0.06 lb × (16 oz/ 1 lb) = 0.96 oz

6% of a pound = 0.96 oz

0.96 oz < 1 oz

Due to the transitive property, we can affirm that 6% of a pound does not weigh more than 1 ounce.

Anni [7]3 years ago
3 0
6% of 1 pound is 96 ounces so yes.
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Answer:

  about 616 in²

Step-by-step explanation:

The area of a sphere is given by ...

  A = πd²

For a ball 14 inches in diameter, the area is ...

  A = π(14 in)² = 196π in²

  A ≈ 616 in²

The area of the required plastic is about 616 square inches.

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Reduce to simplest form.<br> -3/2 - 3/8
svetlana [45]

Answer:

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3 years ago
ANSWER THE QUESTIONS PLEASE IT IS ON BEARING
artcher [175]

9514 1404 393

Answer:

  (i) x° = 70°, y° = 20°

  (ii) ∠BAC ≈ 50.2°

  (iii) 120

  (iv) 300

Step-by-step explanation:

(i) Angle x° is congruent with the one marked 70°, as they are "alternate interior angles" with respect to the parallel north-south lines and transversal AB.

  x = 70

The angle marked y° is the supplement to the one marked 160°.

  y = 20

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(ii) The triangle interior angle at B is x° +y° = 70° +20° = 90°, so triangle ABC is a right triangle. With respect to angle BAC, side BA is adjacent, and side BC is opposite. Then ...

  tan(∠BAC) = BC/BA = 120/100 = 1.2

  ∠BAC = arctan(1.2) ≈ 50.2°

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(iii) The bearing of C from A is the sum of the bearing of B from A and angle BAC.

  bearing of C = 70° +50.2° = 120.2°

The three-digit bearing of C from A is 120.

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(iv) The bearing of A from C is 180 added to the bearing of C from A:

  120 +180 = 300

The three-digit bearing of A from C is 300.

8 0
3 years ago
Evaluate the following expression:
Maurinko [17]

Answer:

1

Step-by-step explanation:

{6}^{2} - 7(8 + 6 -  {3}^{2})

Using BODMAS we'll have to solve what is in the bracket first

=  {6}^{2}  - 7(14 - 9) \\  =  {6}^{2}  - 7(5) \\

Multply -7 by 5 still using the rule of BODMAS

=  {6}^{2}  - 35 \\  = 36 - 35 \\  = 1

<h2><em>OladipoSeun</em><em>♡˖꒰ᵕ༚ᵕ⑅꒱</em></h2>
5 0
3 years ago
The distribution of the number of siblings for students at a large high school is skewed to the right with mean 1.8 siblings and
morpeh [17]

Answer:

E. Approximately normal with standard deviation less than 0.7 sibling

Step-by-step explanation:

To solve this question, we use the Central Limit theorem.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

Skewed right distribution, with \mu = 1.8, \sigma = 0.7

Sampling distribution of the sample mean for samples of size 100

By the Central Limit Theorem, they will be approximately normal, with mean \mu = 1.8, and standard deviation s = \frac{0.7}{\sqrt{100}} = 0.07

So the correct answer is:

E. Approximately normal with standard deviation less than 0.7 sibling

4 0
3 years ago
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