All categories

2 years ago

What single decimal multiplier would you use to increase by 15% followed by a 14% increase?

Mathematics

1 answer:

Yuliya22 [10]2 years ago

3 0

Answer:

The single decimal multiplier that I will use to increase by 15% followed by a 14% increase is 1.311

Step-by-step explanation:

Let

x ----> the number

Remember that

$100\%+15\%=115\%=115/100=1.15$

$100\%+14\%=114\%=114/100=1.14$

we know that

Increase a number by 15% followed by a 14% increase is the same that multiply the number for (1.15)(1.14)

$x(1.15)(1.14)=1.311x$

therefore

The single decimal multiplier that I will use to increase by 15% followed by a 14% increase is 1.311

You might be interested in

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

Of 360 consumers polled some purchased ice cream some purchase frozen yogurt and some purchased both if 260 people polled purcha

Xelga [282]

Total customers are = 360

Number of polled people who purchased ice cream = 260

Number of polled people who purchased both ice cream and yogurt = 250

Hence, people who purchased only ice cream are= $260-250=10$

And $360-10=350$ are people who purchased yogurt.

So, people who purchased yogurt only = $360-(250+10)=100$

8 0

1 year ago

silvia compro la misma licuadora que daniela, que costaba $355.5, pero como abonó con tarjeta de credito, le recargaron un 16%,

Debora [2.8K]

Answer:

Costo final= $412.38

Step-by-step explanation:

Dada la siguiente información:

Costo inicial= $355.5

Recargo de la tarjeta= 16% = 0.16

<u>Para calcular el costo final que debe pagar Silvia, debemos usar la siguiente información:</u>

Costo final= costo inicial*(1 + recargo)

Costo final= 355.5*1.16

Costo final= $412.38

4 0

1 year ago

Using the quadratic formula to solve 11x2 – 4x = 1, what are the values of x? StartFraction 2 Over 11 EndFraction plus-or-minus

slamgirl [31]

Answer:

$x=\frac{2}{11}\pm\frac{\sqrt{15}} {11}$

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$11x^{2} -4x=1$

equate to zero

$11x^{2} -4x-1=0$

$a=11\\b=-4\\c=-1$

substitute in the formula

$x=\frac{-(-4)\pm\sqrt{-4^{2}-4(11)(-1)}} {2(11)}$

$x=\frac{4\pm\sqrt{60}} {22}$

$x=\frac{4\pm2\sqrt{15}} {22}$

$x=\frac{2\pm\sqrt{15}} {11}$

$x=\frac{2}{11}\pm\frac{\sqrt{15}} {11}$

therefore

StartFraction 2 Over 11 EndFraction plus-or-minus StartFraction StartRoot 15 EndRoot Over 11 EndFraction

7 0

2 years ago

Read 2 more answers

. Cara is playing a game. For each dart she can toss onto a board, she

Kaylis [27]

She will need 12 seconds because -20•12=-240

6 0

2 years ago