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Step2247
1 month ago
12

AREA, PERIMETER & VOLUME QUESTION

Mathematics
1 answer:
Inessa [12.5K]1 month ago
7 0
1. Selected Case B. 2. 9 cm³. 3. 20 cm. 4. 4.5 m³. Explanation: In question 1, we need to fit a drum with a volume of 14,000 cm³. The volume of a cylinder can be calculated via the formula πr²h. For Case A, with r = 100 mm (10 cm) and h = 300 mm (30 cm), the total volume is approximately 9424.78 cm³, insufficient for the given drum. Case B, with r = 200 mm (20 cm) and h = 30 cm, gives a volume of approximately 37699.11 cm³. Case C with r = 32 cm and h = 250 mm (25 cm) results in a volume of about 80424.77 cm³. The smallest volume among Cases B and C is Case B at 37699.11 cm³, thus it is the correct choice. For question 2, the dimensions of the speaker are Length = 45 cm = 0.45 m, Width = 0.4 m, Height = 50 cm = 0.5 m, leading to a volume of 0.09 m³ or 9 cm³. Question 3 involves a speaker with a volume of 30,000 cm³ with Length = 30 cm = 0.45 m and Height = 500 mm = 50 cm, requiring to find its Width: 30,000 = 30 × W × 50, hence W = 20 cm. For question 4, with dimensions of Base = 2 m, Length = 3 m, Height =1.5 m, the volume of the prism is calculated as 4.5 m³.
You might be interested in
If 6 components are drawn at random from the container, the probability that at least 4 are not defective is . If 8 components a
Svet_ta [12734]
The question is missing some information. It should be phrased as follows:

<span><span>A container has 50 electronic components, with 10 identified as defective. If 6 components are randomly selected from the container, what is the probability that at least 4 of them are not defective? Additionally, if 8 components are drawn at random from the container, what is the probability that exactly 3 are defective?

</span>Answers
<span>Part 1.  0.02
Part 2. </span></span>0.0375<span><span>

</span>Explanation
Probability denotes the likelihood of an event occurring. It is computed as:
probability = (Number of favorable outcomes)/(Number of total outcomes)

Part 1
When 6 components are chosen, if 4 are confirmed functioning, then 2 must be defective.
P(at least 4 functional) = 4/40</span>× 2/10
                                            = 1/10 × 1/5
                                            = 1/50
                                            = 0.02

Part 2
Choosing 8 components, if 3 are defective, then 5 are functioning.
P(3 defective) = 3/40 × 5/10
                             = 15/400
                             = 3/80
                            = 0.0375
4 0
2 months ago
A 95% confidence interval for the population proportion of professional tennis players who earn more than 2 million dollars a ye
Svet_ta [12734]

Response:

e. 545

Detailed explanation:

In a survey sample containing n individuals, with a success probability of \pi, and a confidence level of 1-\alpha, the ensuing confidence interval for proportions is established.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

Wherein

z denotes the z-score corresponding to a probability value of 1 - \frac{\alpha}{2}.

For this scenario, we find:

The estimate averages the two bounds. Thus \pi = \frac{0.82+0.88}{2} = 0.85

95% confidence level

Consequently, z represents the z value corresponding to the p-value of 1 - \frac{0.05}{2} = 0.975, hence Z = 1.96.

The lower boundary of this interval is:

L = \pi - z\sqrt{\frac{\pi(1-\pi)}{n}}

In this query, L = 0.82. Therefore

0.82 = 0.85 - 1.96\sqrt{\frac{0.85*0.15}{n}}

1.96\sqrt{\frac{0.85*0.15}{n}} = 0.03

0.03\sqrt{n} = 1.96\sqrt{0.85*0.15}

\sqrt{n} = \frac{1.96\sqrt{0.85*0.15}}{0.03}

(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.85*0.15}}{0.03})^{2}

n = 544.23

Thus, the accurate response is:

e. 545

4 0
1 month ago
Every week, Enrique walks one mile each day for 4 days of the week. He can usually walk one mile in about 14-16 minutes. Estimat
Svet_ta [12734]
The estimated distance is 480 miles.
4 0
2 months ago
Kenny is thinking of two numbers greater than 10.
Leona [12618]

Response:

The numbers are 21 and 28

Step-by-step clarification:

To identify the two numbers from the information provided in the question

Firstly, the two numbers need to be greater than 10

The highest common factor for both is 7

The least common multiple is 84

As 7 divides both, the numbers must be multiples of 7

Let’s list the multiples of 7 that are more than 10

14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84,

The numbers cannot exceed 84

Among these, the only numbers that could potentially be 21 and 28

The rationale for this conclusion is that they fulfill the required conditions

8 0
1 month ago
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