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2 years ago

Is 110,571 divisible by 3?

Mathematics

1 answer:

MAXImum [283]2 years ago

3 0

Answer:

yes, it is.

Step-by-step explanation:

110,571 divided by 3 is 36857. So, since the answer is not a decimal or a fraction, 110,571 is divisable by 3

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How do you solve for k G=3 pi k

Paul [167]

Answer: k=G/ 3 pi

Step-by-step explanation:just make k the subject of the formulae by dividing through by 3 pi

4 0

3 years ago

Please help STRUGGLING with math few math questions! Ty

-BARSIC- [3]

Question 1:

F(x) and g(x) are like variables, just plug into the equation.

f(x) + g(x) = (x + 6) + (12x - 7)

x+6+12x-7 = 13x-1

Question 2: f(3) + g(-1)

You plug in the x-values into the equation, and then take the answer and add them together.

f(3) = 3+4

g(-1) = 12(-1)-6

f(3) = 7

g(-1) = -18

7 + (-18) = -11

Question 3:

This is similar to question 1, plug in the variables and simplify.

9x - (7x+3)

Remember to distribute the "-"

9x - 7x - 3

2x - 3

6 0

3 years ago

Read 2 more answers

Find the value of y, rounded to the nearest tenth. Please help me, I'd appreciate it!!

Blababa [14]

If a secant and a tangent of a circle are drawn from a point outside the circle, then the product of the lengths of the secant and its external segment equals the square of the length of the tangent segment.

y² = 7(15+7)
y² = 7*22
y² = 154
y = √154
y = 12.4 ← to the nearest tenth

8 0

2 years ago

Mrs Hong had some milk. She used of it to bake some waffles and

Sergio [31]

Answer:

Mrs. Hong originally had 900 ml of milk

Explanation:

The complete question is in the attached image

Assume that the amount of milk Mrs. Hong originally had was x ml

We are given that:

1- She used $\frac{1}{6}$ of it to bake the waffles, therefore:

Amount used in baking waffles = $\frac{1}{6}x$ ml

2- She used 300 ml to bake a pie

3- Amount left is half the original amount, therefore:

Amount left = $\frac{1}{2}x$ ml

We know that:

Total amount = amount used in waffles + amount used in pie + amount left

Substitute with the givens and solve for x:

Total amount = amount used in waffles + amount used in pie + amount left

$x = \frac{1}{6}x + 300 + \frac{1}{2}x\\ \\x-\frac{1}{6}x-\frac{1}{2}x = 300\\ \\\frac{1}{3}x = 300\\ \\x = 300*3 = 900 ml$

This means that:

She originally had 900 ml of milk

Hope this helps :)

7 0

2 years ago

A cellular phone company monitors monthly phone usage. The following data represent the monthly phone use in minutes of one part

Sergeu [11.5K]

Answer:

The standard deviation increased but there was no change in the interquantile range

Step-by-step explanation:

We are given the following data in the question:

320, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428.

n = 20

a) Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{8726}{20} = 436.3$

Sum of squares of differences =

13525.69 + 640.09 + 7796.89 + 10140.49 + 265.69 + 161.29 + 660.49 + 1108.89 + 313.29 + 6512.49 + 6193.69 + 6130.89 + 2.89 + 10962.09 + 2430.49 + 4664.89 + 4316.49 + 0.49 + 28.09 + 68.89 = 75924.2

$S.D = \sqrt{\frac{75924.2}{19}} = 63.21$

Sorted Data = 320, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

$IQR = Q_3 - Q_1\\Q_3 = \text{upper median},\\Q_1 = \text{ lower median}$

$Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}$

$Median = \frac{431 + 437}{2} = 434$

$Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482$

IQR = $482 - 395 = 87$

b) After changing the observation

0, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428

$Mean =\displaystyle\frac{8406}{20} = 420.3$

Sum of squares of differences =

176652.09 + 86.49 + 5227.29 + 13618.89 + 0.09 + 823.69 + 1738.89 + 299.29 + 1135.69 + 9350.89 + 8968.09 + 3881.29 + 313.29 + 14568.49 + 1108.89 + 2735.29 + 6674.89 + 278.89 + 114.49 + 59.29 = 247636.2

$S.D = \sqrt{\frac{247636.2}{19}} = 114.16$

Sorted Data = 0, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

$Median = \frac{431 + 437}{2} = 434$

$Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482$

IQR = $482 - 395 = 87$

Thus. the standard deviation increased but there was no change in the interquantile range.

5 0

3 years ago