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Jet001 [13]
3 years ago
9

Whats the perfect square of 63 rounding it to the hundreths place

Mathematics
1 answer:
Leni [432]3 years ago
4 0
The answer to your question is 7.94 that the perfect square root of 63
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What is the product of 1/3 and 4/7?
Travka [436]

Answer:

4/21

Step-by-step explanation:

Multiply 1 and 4. Then multiply 3 and 7. So 1x4=4, and 3x7=21. Therefore, the answer is 4/21. In multiplying fractions, you just multiply straight across (: (Hope this helps)

7 0
2 years ago
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A racing car used 255 litres of fuel to complete a 340 km race. How many litres of fuel did the car use every 100km?
natka813 [3]
340 = 255
100 = ?

100 times 255 = 25500
25500 divided by 340 is 75
75 ltrs
7 0
3 years ago
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What is the Multiples of 8?
yan [13]

Answer: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 136, 144, 152, 160, 168, 176, 184, 192, 200, 208, 216, 224, 232, 240, 248, 256, 264, 272, 280, 288, 296, 304, 312, 320, 328, 336, 344, 352, 360, 368, 376, 384, 392

Step-by-step explanation:

Basically just count by 8. Simple.

4 0
3 years ago
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N independent random samples of 10 men and 10 women in a coed basketball league, the numbers of points scored per season are giv
faltersainse [42]

Answer:

Step-by-step explanation:

Hello!

a)

To compare both datasets I've made a box plot for them.

The yellow one belongs to the points scored by the men and the green one is for the points scored by the women.

Women: The box seems symmetrical, the median and the mean (black square) are almost the same. The whiskers are almost the same length, you could say that in general the points scored by the women have a symmetrical distribution.

Men: The box looks a little skewed to the right, the median is closer to the 1st quantile and the mean is greater than the median. The whiskers are the same length, in general the distribution seems almost symmetrical.

b)

As you can see in the box plot  the median and mean of the points scored by the women are almost the same. Using the data I've calculated both values:

Me= 24.50

X[bar]= 23.80

The median divides the sample in halves (50-50),  it shows you where the middle of the distribution is.

The average shows you the value that centers the distribution, meaning, it is the value around which you'll find most of the data set. It summarizes best the sample information and therefore is a better measure of center.

c.

As you can see in the box plots, there are no outliers in both distributions.

An outlier is an observation that is significantly distant from the rest of the data set. They usually represent experimental errors (such as a measurement) or atypical observations. Some statistical measurements, such as the sample mean, are severely affected by this type of values and their presence tends to cause misleading results on a statistical analysis.

Considering the 1st quartile (Q₁), the 3rd quartile (Q₃) and the interquartile range IQR, any value X is considered an outlier if:

X < Q₁ - 1.5 IQR

X > Q₃ + 1.5 IQR

Or extreme outliers if:

X < Q₁ - 3 IQR

X > Q₃ + 3 IQR

For the women data set:

Q₁= 17; Q₃= 30; IQR= 30 - 17= 13

Lower outliers: X < Q₁ - 1.5 IQR= 17-1.5*13= -2.5 ⇒ The minimum value recorded is 03, so there are no outliers.

Upper outliers: X > Q₃ + 3 IQR= 30 + 1.5*13= 49.5 ⇒ The maximum value registered is 41, so there are no outliers.

I hope this helps!

6 0
2 years ago
What is the solution to -x2 + 5x - 3
CaHeK987 [17]

The solution to the equation is x=\frac{5-\sqrt{13}}{2} and x=\frac{5+\sqrt{13}}{2}

Explanation:

Given that the equation is -x^2+5x-3=0

We need to determine the solution of the equation.

The solution of the equation can be determined using the quadratic formula.

The quadratic formula is given by

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

Hence, from the equation, we have,

a=-1,\:b=5,\:c=-3

Substituting these values in the quadratic formula, we get,

x=\frac{-5\pm \sqrt{5^2-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}

Simplifying, we get,

x=\frac{-5\pm \sqrt{25-12}}{-2}

Simplifying the terms within the root, we get,

x=\frac{-5\pm \sqrt{13}}{-2}

Thus, the roots of the equation are x=\frac{-5+ \sqrt{13}}{-2} and x=\frac{-5- \sqrt{13}}{-2}

Taking out the negative sign, we get,

x=\frac{-(5- \sqrt{13})}{-2} and x=\frac{-(5+ \sqrt{13})}{-2}

Cancelling the negative sign, we get,

x=\frac{5-\sqrt{13}}{2} and x=\frac{5+\sqrt{13}}{2}

Thus, the solutions of the equation are x=\frac{5-\sqrt{13}}{2} and x=\frac{5+\sqrt{13}}{2}

7 0
3 years ago
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