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11111nata11111 [884]
4 years ago
10

Given a square with area x, you can use the formula d = 1.4x 1 2 to estimate the length of the diagonal of the square. Use the f

ormula to estimate the length of the diagonal of a square with area 100 cm2 . Show your work!!!
Mathematics
1 answer:
Firlakuza [10]4 years ago
5 0

Answer:

<h2>14cm</h2>

Step-by-step explanation:

Given a length of the diagonal of a square modeled by the equation

d = 1.4x^1/2 where x is the area of the square, to find the estimate of the length of the square if area is 100cm², we will substitute x = 100 into the modeled equation to get s as shown;

d = 1.4 x^{1/2}\\\\d = 1.4 (100)^{1/2}\\\\d = 1.4\sqrt{100}\\\\d = 1.4 * 10\\\\d = \frac{14}{10} * 10\\ \\d = 14

<em>Hence the length of the diagonal of the square using the modeled equation is 14cm</em>

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Greater than 7 tens and less than 8 tens. The number has 4 ones
prohojiy [21]
Answer is 74. Greater than 70 less than 80 so some where in the 70s. It has 4 ones so it would be 74.
6 0
3 years ago
How do you solve this? <br>Y x 49/36=7/12
Triss [41]
Divide both sides by 49/36 to get y by itself:

\frac{7}{12} \div \frac{49}{36}

\frac{7}{12} \times \frac{36}{49} = \frac{254}{588}

Simplify this fraction:

\frac{254}{588} \div \frac{2}{2} = \frac{127}{294}

Now you have your answer:

y = \frac{127}{294}


5 0
3 years ago
Explain the circumstances for which the interquartile range is the preferred measure of dispersion. What is an advantage that th
KatRina [158]

Answer:

Explain the circumstances for which the interquartile range is the preferred measure of dispersion

Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.

What is an advantage that the standard deviation has over the interquartile​ range?

The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.

Step-by-step explanation:

Previous concepts

The interquartile range is defined as the difference between the upper quartile and the first quartile and is a measure of dispersion for a dataset.

IQR= Q_3 -Q_1

The standard deviation is a measure of dispersion obatined from the sample variance and is given by:

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

Solution to the problem

Explain the circumstances for which the interquartile range is the preferred measure of dispersion

Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.

What is an advantage that the standard deviation has over the interquartile​ range?

The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.

8 0
3 years ago
A pyramid of volume 344 square units is sliced into two pieces by a plane parallel to the base of the pyramid and 10 units away
Tom [10]

Answer:

80 units

Step-by-step explanation:

v = 1/3 bh

v₁ = 1/3 * b * h₁ = 344

v₂ = 1/3 * b * h₂ = 301

v₁ / v₂ = h₁ / h₂ = 344 / 301

h₂ = h₁ - 10

h₁ / (h₁ - 10) = 344 / 301

344 * (h₁ - 10) = 301 * h₁

344* h₁ - 3440 = 301 * h₁

43 * h₁ = 3440

h₁ = 80

4 0
3 years ago
Can you guys help me? The questions is... A soccer team eats lunch at a restaurant on the way to a tournament total cost of the
mr Goodwill [35]

Answer: The tip left for the servers was  $14.80

Step-by-step explanation:

Hi, to answer this question we have to analyze the information given:

  • <em>Total cost of the food (before tax) = $82.25 </em>
  • <em>Tip = 18% of the amount before the tax </em>

So, to calculate the tip we have to multiply the cost of the food before tax ($82.25) by the 0.18 (percentage in decimal form, 18/100= 0.18).

Mathematically speaking:

82.25 x 0.18 = $14.80

The tip left for the servers was $14.80

3 0
3 years ago
Read 2 more answers
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