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3 years ago

11

What is B-CF?

1 answer:

Sholpan [36]3 years ago

7 0

I'm guessing on the make up of the matrices.
First off let's look at [C][F].
[C]=
[F]=
[C][F]=
where each element of [C][F] comes from multiplying a row of [C] with a column of [F].
Example: First element is product of first row and first column.

.
.
.
Now that we have [C][F], we can subtract it from [B], element by element,

[B]-[C][F]=
[B]-[C][F]=
.
.
.
If this is not how the matrices look,please re-state the problem and be more specific about the make up of the matrices (rows x columns).
Here's an example.
[A] is a 2x2 matrix. A=[1,2,3,4].
The assumption is that [A] looks like this,
[A]=
[B] is a 3x2 matrix. B=[5,6,7,8,9,10]
[B]=

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The amount of soda dispensing machine pours into a can of soda is normally distributedwith a mean of 12 ounces and a standard de

kaheart [24]

Answer:

0.1056

Step-by-step explanation:

Mean(μ) = 12 ounce

Standard deviation (σ) = 0.2 ounce

P(x<11.75) = ???

Let x be the random variable for the amount of soda a machine will dispense.

Using normal distribution

Z = (x - μ) / σ

Z = (11.75 - 12) / 0.2

Z = (-0.25)/0.2

Z = -1.25

From the normal distribution table

1.28 is 0.3944

Φ(z) is the tabular value of z

Φ(z) = 0.3944

Recall that when Z is negative

P(x<a) = 0.5 - Φ(z)

P(x < 11.75) = 0.5 - 0.3944

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7 0

3 years ago

In 1859, 24 rabbits were released into the wild in Australia, where they had no natural predators. Their population grew exponen

Mashcka [7]

Answer:

a) P' = P

$P(t) = 24e^{0.693t}$ where t is step of 6 months

b) 7.7 years

c)1064.67 rabbits/year

Step-by-step explanation:

The differential equation describing the population growth is

$\frac{dP}{dt} = P$

Where t is the range of 6 months, or half of a year.

P(t) would have the form of

$P(t) = P_0e^{kt}$

where $P_0 = 24$ is the initial population

After 6 month (t = 1), the population is doubled to 48

$P(1) = 24e^k = 48$

$e^k = 2$

$k = ln(2) = 0.693$

Therefore $P(t) = 24e^{0.693t}$

where t is step of 6 months

b. We can solve for t to get how long it takes to get to a population of 1,000,000:

$24e^{0.693t} = 1000000$

$e^{0.693t} = 1000000 / 24 = 41667$

$0.693t = ln(41667) = 10.64$

$t = 10.64 / 0.693 = 15.35$

So it would take 15.35 * 0.5 = 7.7 years to reach 1000000

c. $P' = P_0ke^{kt}$

We need to resolve for k if t is in the range of 1 year. In half of a year (t = 0.5), the population is 48

$24e^{0.5k) = 48$

$0.5k = ln2 = 0.693$

$k = 1.386$

Therefore, $P' = 1.386*24e^{1.386t}$

At the mid of the 3rd year, where t = 2.5, we can calculate P'

$P' = 1.386*24e^{1.386*2.5} = 1064.67$ rabbits/year

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3 years ago

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Studentka2010 [4]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago

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