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3 years ago

I am 10 what is 150% of me

Mathematics

2 answers:

Alla [95]3 years ago

8 0

150 percent of 10 is 15
the correct answer is 15

Mandarinka [93]3 years ago

5 0

100% + 50% of something equals 150% of something.

100% of 10 is 10
50% of 10 is 5
10+5=
15

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Can someone help me solve this?

svetoff [14.1K]

Answer:

honestly they all are the same i had a similar question and thats what i put so yeah sorry if i didn`t help

Step-by-step explanation:

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2 years ago

Given the sample −4, −10, −16, 8, −12, add one more sample value that will make the mean equal to 3.

Agata [3.3K]

Answer: The value to add to the set is 52

=======================================================

For a moment, ignore the extra value we'll add in later.

The given set of values add to -34.

Let x be the extra value we add in. It adds to the -34 to get -34+x, then we divide by 6 because the set {-4,-10,-16,8,-12,x} has 6 items.

So we get the mean (-34+x)/6

Set this equal to the mean we want (3) and solve for x

(-34+x)/6 = 3

-34+x = 6*3

-34+x = 18

x = 18+34

x = 52

The set {-4,-10,-16,8,-12,x} updates to {-4,-10,-16,8,-12,52} and this set has a mean of 3.

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2 years ago

What is 6.5555555555556 in a fraction

shusha [124]

Answer:

59/9

explanation: just divide 59 divided by 9

6 0

2 years ago

Read 2 more answers

I need help on this question

Ainat [17]

Answer choice 3 is correct
The answer is 3x+18=147°, because a central angle is equal to the amount of degrees of its arc.

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3 years ago

Suppose x is a real number and epsilon > 0. Prove that (x - epsilon, x epsilon) is a neighborhood of each of its members; in

kaheart [24]

Answer:

See proof below

Step-by-step explanation:

We will use properties of inequalities during the proof.

Let $y\in (x-\epsilon,x+\epsilon)$ . then we have that $x-\epsilon$ . Hence, it makes sense to define the positive number delta as $\delta=\min\{x+\epsilon-y,y-(x-\epsilon)\}$ (the inequality guarantees that these numbers are positive).

Intuitively, delta is the shortest distance from y to the endpoints of the interval. Now, we claim that $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ , and if we prove this, we are done. To prove it, let $z\in (y-\delta,y+\delta)$ , then $y-\delta$ . First, $\delta \leq y-(x-\epsilon)$ then $-\delta \geq -y+x-\epsilon$ hence $z>y-\delta \geq x-\epsilon$

On the other hand, $\delta \leq x+\epsilon-y$ then $z$ hence $z$ . Combining the inequalities, we have that $x-\epsilon$ , therefore $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ as required.

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3 years ago