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4 years ago

Radius OB measures x units. Which expression represents

Mathematics

1 answer:

777dan777 [17]4 years ago

7 0

Answer:

Circumference of the smaller circle = 2.1x units

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Two similar circles are shown. The circumference of the larger circle, with radius OB, is 3 times the circumference of the smaller circle, with radius OA. Radius OB measures x units. Which expression represents the circumference of the smaller circle with radius OA ?

Radius of the larger circle OB = x units

Radius of the smaller circle OA = y units

Circumference of larger circle = 2πx units

Circumference of smaller circle = 2πy units

Since circumference of larger circle is 3 times the circumference of smaller circle,

2πx = 3(2πy)

2πx = 6πy

x = 3y

y = $\frac{x}{3}$

Therefore, circumference of the smaller circle = $2\pi (\frac{x}{3})$ units

= $\frac{2\pi x}{3}$ units

= 2.094x

≈ 2.1x

Circumference of the smaller circle circle = 2.1x units

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Answer:

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Step-by-step explanation:

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<em>Greetings from Brasil</em>

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4 years ago

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Free_Kalibri [48]

Answer:

Step-by-step explanation:

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3 years ago

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Step-by-step explanation:

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Allied Corporation is trying to sell its new machines to Ajax. Allied claims that the machine will pay for itself since the time

kvasek [131]

Answer:

z(s) is in the acceptance region. We accept H₀ we did not find a significantly difference in the performance of the two machines therefore we suggest not to buy a new machine

Step-by-step explanation:

We must evaluate the differences of the means of the two machines, to do so, we will assume a CI of 95%, and as the interest is to find out if the new machine has better performance ( machine has a bigger efficiency or the new machine produces more units per unit of time than the old one) the test will be a one tail-test (to the left).

New machine

Sample mean x₁ = 25

Sample variance s₁ = 27

Sample size n₁ = 45

Old machine

Sample mean x₂ = 23

Sample variance s₂ = 7,56

Sample size n₂ = 36

Test Hypothesis:

Null hypothesis H₀ x₂ - x₁ = d = 0

Alternative hypothesis Hₐ x₂ - x₁ < 0

CI = 90 % ⇒ α = 10 % α = 0,1 z(c) = - 1,28

To calculate z(s)

z(s) = ( x₂ - x₁ ) / √s₁² / n₁ + s₂² / n₂

s₁ = 27 ⇒ s₁² = 729

n₁ = 45 ⇒ s₁² / n₁ = 16,2

s₂ = 7,56 ⇒ s₂² = 57,15

n₂ = 36 ⇒ s₂² / n₂ = 1,5876

√s₁² / n₁ + s₂² / n₂ = √ 16,2 + 1.5876 = 4,2175

z(s) = (23 - 25 )/4,2175

z(s) = - 0,4742

Comparing z(s) and z(c)

|z(s)| < | z(c)|

z(s) is in the acceptance region. We accept H₀ we did not find a significantly difference in the performance of the two machines therefore we suggest not to buy a new machine

The very hight dispersion of values s₁ = 27 is evidence of frecuent values quite far from the mean

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3 years ago