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podryga [215]
3 years ago
14

The sum of 367 and 863 is equal to the sum of 863 and another number. what is that number?

Mathematics
2 answers:
dangina [55]3 years ago
7 0

Answer:

Sum of 367 and 863 is equal to the sum of 863 and 367. Thus, 367 is the required number.

Step-by-step explanation:

<u>Commutative property of Addition</u>

Let a and b be two real numbers. Then commutative property of addition states that the sum of a and b is equal to the sum of b and a that is

a + b = b + a

Similarly, we a re given here,

367 + 863 = 1230\\1230 = 863 + 367

Thus, sum of 367 and 863 is equal to the sum of 863 and 367.

Thus, 367 is the required number.

oksian1 [2.3K]3 years ago
4 0
The other no is
863-367 =496
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If the inspection division of a county weights and measures department wants to estimate the mean amount of soft-drink fill in 2
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Answer:

n=97

Step-by-step explanation:

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Assuming the X follows a normal distribution

X \sim N(\mu, \sigma=0.01)

We know that the margin of error for a confidence interval is given by:

Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

The next step would be find the value of \z_{\alpha/2}, \alpha=1-0.95=.05 and \alpha/2=0.025

Using the normal standard table, excel or a calculator we see that:

z_{\alpha/2}=1.96

If we solve for n from formula (1) we got:

\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}

n=(\frac{z_{\alpha/2} \sigma}{Me})^2

And we have everything to replace into the formula:

n=(\frac{1.96(0.05)}{0.01})^2 =96.04

And if we round up the answer we see that the value of n to ensure the margin of error required \pm=0.01 Liters is n=97.

5 0
4 years ago
A pool measuring 10 meters by 26 meters is surrounded by a path of uniform​ width, as shown in the figure. If the area of the po
igomit [66]

Based on the measurements of the pool and the path of uniform width, the area of the width and length of the path can be found to be 28.8 meters.

<h3>How to find the width and length?</h3>

First, find the area of the pool:

= Length x Width

= 10 x 26

= 260 meters ²

The area of the path can therefore be found to be:

= Total area of pool and path combined - Area of pool

= 1,092 - 260

= 832 meters²

Seeing as the path has uniform width, that means that the width is the same and the length so the width and length of the pool is:
= √area of the path

= √832

= 28.8

In conclusion, the width and length of the path are 28.8 meters.

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8 0
1 year ago
1. Multiply (x+7)(x-7)
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7 0
3 years ago
Y=3x-2 use a graph to estimate the value of x when y = 3
rodikova [14]

Answer:

5/3

Step-by-step explanation:

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2 years ago
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Suppose in a class of 60 students 5 have no siblings, 26 have one sibling, 14 have two siblings, and 15 have three siblings. Cal
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The relative frequency of students who have three siblings is 25% given that there are 60 students in a class in which 5 have no siblings, 26 have one sibling, 14 have two siblings and 15 have three siblings. This can be obtained by using the formula for relative frequency.

<h3>What is the relative frequency of students who have three siblings?</h3>

Given that,

total number of students in the class = 60

number of students who have no sibling = 5

number of students who have 1 sibling = 26

number of students who have 2 sibling = 14

number of students who have 3 sibling = 15

Formula for relative frequency = f/n, where f is the number of times the data occurred, n is the total number of frequencies.

Therefore,

relative frequency of students who have three siblings = 15/60 = 0.25

In percentage ⇒ 0.25 × 100 = 25%

Hence the relative frequency of students who have three siblings is 25% given that there are 60 students in a class in which 5 have no siblings, 26 have one sibling, 14 have two siblings and 15 have three siblings.

Learn more about relative frequency here:

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2 years ago
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