What is the mode of the following numbers? 67, 76, 67, 76, 67, 23, 32, 23, 32 0

Define x as the first mile marker, thus x + 1 corresponds to the second mile marker. This can be expressed as x + (x + 1) = 561, which simplifies to 2x + 1 = 561. Therefore, 2x equals 560, leading to x being 280. Consequently, the first mile marker is 280, and the second mile marker is 281.

Answer:

A) 20+0.6

B) Domain is [0, 50] (inclusive)

C) 8.33 litres

Step-by-step explanation:

Assuming there is a 40% acid concentration in a 50-litre container.

The quantity of acid in this container amounts to 40% of 50 litres.

Quantity of acid =

litres are extracted.

The acid volume removed equals 40% of litre.

Remaining acid in the container thus becomes = (20 - 40% of ) litre

This is then substituted with pure acid.

Final acid content in the container is therefore = (20 - 40% of + 100% of ) litre

Quantity of acid in the final mixture:

Answer A) Final mixture acid content = 20+0.6

Answer B) cannot exceed 50 litres (the container's initial volume) and must be at least 0 litres.

Thus, the domain is [0, 50] (inclusive)

Answer C)

For the final mixture to contain 50% acid.

The acid volume = 50% of 50 litres = 25 litres

By applying the equation:

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To start, draw a line connecting the circle's center to one endpoint of the chord (water surface) and again to the point with maximum depth. This creates a right triangle. The length of the side up to the water-surface measures 5 cm, and the hypotenuse is 7 cm.

<span>Now, perform the following steps: </span>

<span>Calculate angle θ at the right triangle's corner using: cos θ = 5/7 ⇒ θ = cos ˉ¹ (5/7) </span>

<span>Thus, θ is approximately 44.4°, leading to the angle subtended at the circle center by the water surface being roughly 88.8°. </span>

<span>To find the shaded area, subtract the triangular area above the water in your diagram from the area of the sector. </span>

<span>Shaded area ≃ 88.8/360*area of circle - ½*7*7*sin88.8° </span>
<span>= 88.8/360*π*7² - 24.5*sin 88.8° </span>
<span>which is approximately 13.5 cm² </span>
<span>(using area of ∆ = ½.a.b.sin C for the triangle) </span>



<span>b) </span>

<span>The water's volume is equal to the cross-sectional area multiplied by the length </span>
<span>≈ 13.5 * 30 cm³ </span>
<span>≈ 404 cm³</span>

I hope this was helpful

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