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3 years ago

A. 16%

Mathematics

2 answers:

k0ka [10]3 years ago

8 0

I'd say it would be 24% but i may be wrong

netineya [11]3 years ago

6 0

There are a total of 50 data points or entries.
Total A+B+C+D+E+F=50
Total 5+12+8+10+6+9=50
There are 8 C's being picked out of 50.
8 / 50 = 0.16 or 16%

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Which inequality is true?

ICE Princess25 [194]

Answer:

∠6 > ∠3 is TRUE.

Step-by-step explanation:

We know that acute angle is less than 90 degree while obtuse angle is more than 90 degree. By looking at the given angles, we can easily determine which angle is obtuse or acute. then we can decide which angle is bigger to find the answer.

∠10 > ∠7

FALSE;

Because∠10 is acute while ∠7 is obtuse.

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∠6 > ∠3

TRUE;

Because∠3 is acute while ∠6 is obtuse.

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∠9 > ∠11

FALSE;

Because∠9 is acute while ∠11 is obtuse.

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∠8 > ∠11

FALSE;

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2 years ago

Using the factoring process, what constant is added to the RIGHT side of the quadratic equation 2x2? 9x ___ = ?9 ___ in order to

IgorC [24]

Normally you would only solve by completing the square if a is 1 and b is even. But the answer would be half of b. So half of 9 is 4.5 or 4(1/2).

3 0

3 years ago

Read 2 more answers

Suppose that you take 120 mg of an antibiotic every 4 hr. The half-life of the drug is 4 hr (the time it takes for half of the

vodomira [7]

Answer:

The steady state amount of antibiotic in the bloodstream when t --> ∞ is 240 mg.

Step-by-step explanation:

Let the amount of antibiotic in one's bloodstream be given as Aₙ (where n = the number of half lives since the start of usage)

Let's follow the time line of events.

At t = 0 hr, the drug is taken

A₀ = 120 mg

At t = 4 hrs, n = 1, the drug is taken again

A₁ = (0.5×A₀) + 120

A₁ = (0.5×120) + 120 = 180 mg

At t = 8 hrs, n = 2, the drug is taken again,

A₂ = (0.5×A₁) + 120

A₂ = (0.5×180) + 120 = 210 mg

At t = 12 hrs, n = 3, the drug is taken again

A₃ = (0.5×A₂) + 120

A₃ = (0.5×210) + 120 = 225 mg

At this point, it becomes evident that at t = 4n hrs, n = n i.e. n half lives later, the general formula for the amount of the antibiotic in the bloodstream is

Aₙ = 0.5Aₙ₋₁ + 120

where Aₙ₋₁ = The amount of antibiotic in the bloodstream at the time t = 4(n-1) and (n-1) half lives later.

For infinite series, that are increasing in this order, as the value of n --> ∞,

Aₙ = Aₙ₋₁ = K

And our general formula becomes

K = 0.5K + 120

0.5K = 120

K = (120/0.5)

K = 240 mg

Hence, the steady state amount of antibiotic in the bloodstream when t --> ∞ is 240 mg.

Hope this Helps!!!

5 0

3 years ago

Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$