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Sveta_85 [38]
1 year ago
9

Mia is 7.01568 x 10^6 minutes old. Convert her age to more appropriate units using years, months, and days. Assume every other m

onth has 30 days, instead of 31.​
Mathematics
1 answer:
pshichka [43]1 year ago
8 0

Answer:

There are 13 years 6 months and 12 days.

Step-by-step explanation:

Let's convert the scientific notation to standard form of minutes

7.01568x10^6 =7015680 minutes

Now, let's convert this to number of days because each day has 24*60=1440 minutes.

So, number of days in 7015680 minutes is \frac{7015680}{1440}=4872 days

Now, change this 4872 days to years, months and days.

We know 1 year has 12 months and one month has 30 days.

So,

Number of years =\frac{4872} {12*30}=13.53333333 years

Now, convert 0.53333333 years to months.

0.53333333*12=6.4 months.

Now, convert 0.4 months to days.

0.4* 30 =12 days.

So, there are 13 years 6 months and 12 days.

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Let P2 be the vector space of all polynomials of degree 2 or less, and let H be the subspace spanned by 10x2+4xâ1, 3xâ4x2+3, and
lord [1]

I suppose

H=\mathrm{span}\{10x^2+4x-1,3x-4x^2+3,5x^2+x-1\}

The vectors that span H form a basis for P_2 if they are (1) linearly independent and (2) any vector in P_2 can be expressed as a linear combination of those vectors (i.e. they span P_2).

  • Independence:

Compute the Wronskian determinant:

\begin{vmatrix}10x^2+4x-1&3x-4x^2+3&5x^2+x-1\\20x+4&3-8x&10x+1\\20&-8&10\end{vmatrix}=-6\neq0

The determinant is non-zero, so the vectors are linearly independent. For this reason, we also know the dimension of H is 3.

  • Span:

Write an arbitrary vector in P_2 as ax^2+bx+c. Then the given vectors span P_2 if there is always a choice of scalars k_1,k_2,k_3 such that

k_1(10x^2+4x-1)+k_2(3x-4x^2+3)+k_3(5x^2+x-1)=ax^2+bx+c

which is equivalent to the system

\begin{bmatrix}10&-4&5\\4&3&1\\-1&3&-1\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}

The coefficient matrix is non-singular, so it has an inverse. Multiplying both sides by that inverse gives

\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}-\dfrac{6a-11b+19c}3\\\dfrac{3a-5b+2c}3\\\dfrac{15a-26b+46c}3\end{bmatrix}

so the vectors do span P_2.

The vectors comprising H form a basis for it because they are linearly independent.

4 0
2 years ago
Find a linear equation to model this real-world application:
SSSSS [86.1K]

The linear equation to model the company's monthly expenses is y = 2.5x + 3650

<em><u>Solution:</u></em>

Let "x" be the units produced in a month

It costs ABC electronics company $2.50 per unit to produce a part used in a popular brand of desktop computers.

Cost per unit = $ 2.50

The company has monthly operating expenses of $350 for utilities and $3300 for salaries

We have to write the linear equation

The linear equation to model the company's monthly expenses in the form of:

y = mx + b

Cost per unit = $ 2.50

Monthly Expenses = $ 350 for utilities and $ 3300 for salaries

Let "y" be the total monthly expenses per month

Then,

Total expenses = Cost per unit(number of units) + Monthly Expenses

y = 2.50(x) + 350 + 3300\\\\y = 2.5x + 3650

Thus the linear equation to model the company's monthly expenses is y = 2.5x + 3650

3 0
1 year ago
Which shows a recursive and an explicit formula for a sequence whose initial term is 14 and whose common difference is −4? A. A(
liberstina [14]

Answer: A. A(1) = 14; A(n) = (n − 1) −4; A(n) = 14 + (n − 1)(−4)

Step-by-step explanation:

Arithmetic sequence is a sequence that is identified by their common difference. Let a be the first term, n be the number of terms and d be the common difference.

For an arithmetic sequence, common difference 'd' is added to the preceding term to get its succeeding term. For example if a is the first term of a sequence, second term will be a+d, third term will give a+d+d and so on to generate sequence of the form,

a, a+d, a+3d, a+4d...

Notice that each new term keep increasing by a common difference 'd'

The nth term of the sequence Tn will therefore give Tn = a+(n-1)d

If the initial (first) term is 14 and common difference is -4, the nth of the sequence will be gotten by substituting a = 14 and d = -4 in the general formula to give;

Tn = 14+(n-1)-4 (which gives the required answer)

Tn = 14-4n+4

Tn = 18-4n

5 0
1 year ago
Read 2 more answers
What is the volume of a cylinder with base radius 222 and height 666? either enter an exact answer in terms of \piπpi or use 3.1
Leno4ka [110]

In this question, it is given that the base radius is 2 , height is 6 and the value of pi is 3.14 .

And we have to find the volume of the cylinder, and for that, we have to use the following formula

V = \pi r^2 h

Substituting the values of pi,r and h, we will get

V = 3.14(2)^2 * 6 = 75.36 \ cubic \ units

3 0
2 years ago
Read 2 more answers
1) A family consisting of three persons—A, B, and C—goes to a medical clinic that always has a doctor at each of stations 1, 2,
crimeas [40]

Answer:

Step-by-step explanation:

Let us record the station number 1, 2 or 3 for each family member A, B or C.

I am attaching a table containing total outcomes. Outcomes are presented along rows while the assigned station to each member is written along columns. For ease of understanding, 1 3 2 in the table should be interpreted as family member A being assigned to station 1, member B to station 3 and member C to station number 2, respectively.

From table it is clear that the total outcomes possible are 27.

We know that, probability can be defined as,

PROBABITILY = \frac{NUMBER\;OF\;DESIRED\;OUTCOMES}{TOTAL\;NUMBER\;OF\;OUTCOMES}

a) All Members Assigned to the Same Station.

Cases for all members being assigned to same station are as follows:

[1 1 1], [2 2 2], [3 3 3] (outcome number 1, 14 and 27 in the table).

Therefore,

PROBABILITY\;(Case\;a) = \frac{3}{27}\\\\PROBABILITY\;(Case\;a) = 0.111

b) At Most Two Members Assigned to the Same Station.

It means that maximum of 2 members can have the same station. Cases for this situation are as follows:

[1 1 2], [1 1 3], [1 2 1], [1 2 2], [1 3 1], [1 3 3], [2 1 1], [2 1 2], [2 2 1], [2 2 3], [2 3 2],

[2 3 3], [3 1 1], [3 1 3], [3 2 2], [3 2 3], [3 3 1], [3 3 2]

(outcome number 2, 3, 4, 5, 7, 9, 10, 11, 13, 15, 17, 18, 19, 21, 23, 24, 25 and 26 in the table).

Therefore,

PROBABILITY\;(Case\;b) = \frac{18}{27}\\\\PROBABILITY\;(Case\;b) = 0.666

c) All Members Assigned to a Different Station.

For this scenario, we have the following results:

[1 2 3], [1 3 2], [2 1 3], [2 3 1], [3 1 2], [3 2 1] (outcome number 6, 8, 12, 16, 20 and 22 in the table).

Therefore,

PROBABILITY\;(Case\;c) = \frac{6}{27}\\\\PROBABILITY\;(Case\;c) = 0.222

3 0
2 years ago
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