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3 years ago

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Nicholas and Millah are comparing results from a 2-day lab experiment. On Monday, Nicholas’s results showed that 0.917 of the pa

rticles were charged while Millah’s results showed that 85% of the particles were charged. On Tuesday, Nicholas’s results showed that 0.799 of the particles were charged while Millah’s results showed that 93% of the particles were charged. Who on what day had the greatest percentage of charged particles in their experiment?

1 answer:

nalin [4]3 years ago

4 0

Millah Tuesday because they had %93 charged.

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Using the linear equation 4x–3y=12, express: b x in terms of y

Amiraneli [1.4K]

Answer:

y = [4(x-3)]/3

Step-by-step explanation:

4x = 12+3y

3y = 4x-12

y = [4(x-3)]/3

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2 years ago

Whats the word phrase of 49+m?

Arturiano [62]

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3 years ago

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Find the angle between the given vectors. Round your answer, in degrees, to two decimal places. u=⟨2,−6⟩u=⟨2,−6⟩, v=⟨4,−7⟩

NISA [10]

Answer:

$\theta = 108.29$

Step-by-step explanation:

Given

$u =$

$v =$

Required:

Calculate the angle between u and v

The angle $\theta$ is calculated as thus:

$cos\theta = \frac{u.v}{|u|.|v|}$

For a vector

$A =$

$A = a * b$

$cos\theta = \frac{u.v}{|u|.|v|}$ becomes

$cos\theta = \frac{.}{|u|.|v|}$

$cos\theta = \frac{2*6+4*-7}{|u|.|v|}$

$cos\theta = \frac{12-28}{|u|.|v|}$

$cos\theta = \frac{-16}{|u|.|v|}$

For a vector

$A =$

$|A| = \sqrt{a^2 + b^2}$

So;

$|u| = \sqrt{2^2 + 6^2}$

$|u| = \sqrt{4 + 36}$

$|u| = \sqrt{40}$

$|v| = \sqrt{4^2+(-7)^2}$

$|v| = \sqrt{16+49}$

$|v| = \sqrt{65}$

So:

$cos\theta = \frac{-16}{|u|.|v|}$

$cos\theta = \frac{-16}{\sqrt{40}*\sqrt{65}}$

$cos\theta = \frac{-16}{\sqrt{2600}}$

$cos\theta = \frac{-16}{\sqrt{100*26}}$

$cos\theta = \frac{-16}{10\sqrt{26}}$

$cos\theta = \frac{-8}{5\sqrt{26}}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{-8}{5\sqrt{26}})$

$\theta = cos^{-1}(\frac{-8}{5 * 5.0990})$

$\theta = cos^{-1}(\frac{-8}{25.495})$

$\theta = cos^{-1}(-0.31378701706)$

$\theta = 108.288386087$

<em></em> $\theta = 108.29$ <em> (approximated)</em>

4 0

2 years ago

A rabbit population doubles every 4 weeks.There are currently five rabbits in a restricted area. If t represents the time in wee

sertanlavr [38]

Doubling formula is this:

$P(t)=P(2)^{\frac{t}{d}}$
where P=initial number of rabbits
t=time
d=time it takes to doulbe

ok, so 4 weeks is the doubling time so that is 4*7=28 days

we wawnt time=98
and oroiginal number of rabbits is 5 so
$P(98)=5(2)^{\frac{98}{28}}$
$P(98)=5(2)^{3.5}$
$P(98)=5(2^3)(\sqrt{2})$
$P(98)=5(8)\sqrt{2}$
$P(98)=40\sqrt{2}$
so P(98)≈56.56
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about 56 or 57 rabbits in 98 days

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3 years ago

The first three terms of a sequence are given. Round to the nearest thousandth (if necessary).

BabaBlast [244]

Step-by-step explanation:

the introduction of a fraction tells us that we are dealing with multiplications, and therefore a geometric sequence (where every new term is created by multiplying the previous term by a constant factor, the ratio r).

I think your teacher made a mistake, or you made one when typing the question in here.

there is no factor r that creates

15×r = 9

and

9×r = 5/27

it would mean that

15 × r² = 5/27

r² = 5/27 / 15 = 5/27 × 1/15 = 5/405 = 1/81

r = 1/9

but 15 × 1/9 = 5 × 1/3 = 5/3 is NOT 9

and 9 × 1/9 = 9/9 = 1 is NOT 5/27

so, this can't be right.

on the other hand

15 × r = 9

r = 9/15 = 3/5

and then

9 × 3/5 = 27/5

so, either the sequence should have been

15, 5/3, 5/27

or (and I suspect this to be true)

15, 9, 27/5

under that assumption we have

s1 = 15

r = 3/5

sn = sn-1 × r = s1 × r^(n-1) = 15 × (3/5)^(n-1)

s10 = 15 × (3/5)⁹ = 15 × 19683/1953125 =

= 3 × 19683/390625 = 59049/390625 =

= 0.15116544 ≈ 0.151

6 0

2 years ago