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Svetllana [295]
3 years ago
11

Dilution Problem

Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
4 0

Answer:

  76.7 liters

Step-by-step explanation:

You have ...

  C1×V1 = C2×V2

so ...

  V2 = V1×(C1/C2) = (115 L)×(14/21) = 76 2/3 L

  V2 ≈ 76.7 L

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A trapezoidal prism undergoes a dilation. The surface area of the pre-image is 35 m². The surface area of the image is 315 m². T
Westkost [7]

The original object is called pre-image, and its dilated version is called image. The height of the image for the considered case is 17.4 m

<h3>How does dilation affect length, area, and volume of an object?</h3>

Suppose a figure (pre image) is dilated (dilated image) by scale factor of k.

So, if a side of the figure is of length L units, and that of its similar figure is of M units, then:

<u><em>L = k × M</em></u>

where 'k' will be called as scale factor.

The linear things grow linearly like length, height etc.

The quantities which are squares or multiple of linear things twice grow by square of scale factor. Thus, we need to multiply or divide by <u><em>k²</em></u><em> </em>to get each other corresponding quantity from their similar figures' quantities.

<u><em>So </em></u><u><em>area </em></u><u><em>of </em></u><u><em>first figure </em></u><u><em>= k² × area of second figure</em></u>

Similarly,

<u><em>Volume </em></u><u><em>of first figure = k³ × volume of second figure.</em></u>

It is because we will need to multiply 3 linear quantities to get volume, which results in k getting multiplied 3 times, thus, cubed.

For this case, we're given that;

  • Surface area of pre-image = 35 m²
  • Surface area of dilated image = 315 m²
  • Height of pre-image = 5.8 m
  • Height of dilated image.

Let the scale fator of dilation be k.

Then:

<u><em>So area of pre-image = k² × area of dilated image</em></u>

Thus, we get:

35 = k^2 \times 315\\\\k = \sqrt{\dfrac{35}{315}} = 1/3

(took positive root as scaling is done by non-negative factors as only magnitude matters mostly)

Thus, we get:

Height of pre-image = (1/3) × Height of dilated image (say h)

5.8 = (1/3) \times h\\h = 3 \times 5.8 = 17.4 \rm \: m

Thus, the original object is called pre-image, and its dilated version is called image. The height of the image for the considered case is 17.4 m

Learn more about dilation here:

brainly.com/question/3266920

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2 years ago
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Consider a manufacturing process that is producing hypodermic needles that will be used for blood donations. These needles need
Mars2501 [29]

Answer:

\text{Average diameter is 1.65 mm and we decide that it is not 1.65 mm.}                

Step-by-step explanation:

We are given the following in the question:

The needle size should not be too big and too small.

The diameter of the needle should be 1.65 mm.

We design the null and the alternate hypothesis

H_{0}: \mu = 1.65\text{ mm}\\H_A: \mu \neq 1.65\text{ mm}

Sample size, n = 35

Sample mean, \bar{x} = 1.64 mm

Sample standard deviation, s = 0.07 mm

Type I error:

  • It is the error of rejecting the null hypothesis when it is true.
  • It is also known as false positive error.
  • It is the rejecting of a true null hypothesis.

Thus, type I error in this study would mean we reject the null hypothesis that the average diameter is 1.65 mm but actually the average diameters of the needle is 1.65 mm.

Thus, average diameter is 1.65 mm and we decide that it is not 1.65 mm.

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Can someone help me on this math problem PLEASE
egoroff_w [7]

Answer:

QR perpendicular to PT

PR^Q=SR^T=90°

also ANGLES (QPR= STR

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What is the measure of XYZ?<br> A. 75°<br> B. 33<br> C. 54<br> D. 108
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It's D. 108 • ___________
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I need help with this asappp
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The answer would be 45. Thank you very much and good luck on there!
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