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3 years ago

IO.7(4.2)+15.9= 51.25+43.85+14=

Mathematics

1 answer:

katovenus [111]3 years ago

7 0

Answer:

10.7(4.2)+15.9 equals 60.84

51.25+43.85+14 equals 109.1

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The commuting time for a student to travel from home to a college campus is normally distributed with a mean of 30 minutes and a

Rainbow [258]

Answer:

0.1587

Step-by-step explanation:

Let X be the commuting time for the student. We know that $X\sim N(30, (5)^2)$ . Then, the normal probability density function for the random variable X is given by

$f(x) = \frac{1}{\sqrt{2\pi(5)^{2}}}\exp[-\frac{(x-\mu)^{2}}{2(5)^{2}}]$ . We are seeking the probability P(X>35) because the student leaves home at 8:25 A.M., we want to know the probability that the student will arrive at the college campus later than 9 A.M. and between 8:25 A.M. and 9 A.M. there are 35 minutes of difference. So,

$P(X>35) = \int\limits_{35}^{\infty}f(x)dx$ = 0.1587

To find this probability you can use either a table from a book or a programming language. We have used the R statistical programming language an the instruction pnorm(35, mean = 30, sd = 5, lower.tail = F)

6 0

4 years ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

3 years ago

What is the solution of the system of equations? y = –3x + 8 y = –5x – 2

jekas [21]

$y=-3x+8 \\ y=-5x-2 \\ \\ -3x+8=-5x-2 \\ -3x+5x=-2-8 \\ 2x=-10 \\ x=- \frac{10}{2} \\ x=-5 \\ y=-3(-5)+8=15+8=23 \\ \\ \boxed{-5,23}$

6 0

3 years ago

Read 2 more answers

Help please this is due soon!!!!!!!

tensa zangetsu [6.8K]

Answer:

Step-by-step explanation:

Slope of (-4,7),(-7,-15) is 22/3

Slope of the function is 2/11

3 0

3 years ago

Read 2 more answers

On Tuesday the temperature is 1 ºC. By Wednesday it has dropped to –2 ºC. The temperature drops by twice as much from Wednesday

Finger [1]

Answer: $-8^{\circ}C$

Step-by-step explanation:

Given

The temperature on Tuesday is $1^{\circ}C$

Temperature drops to $-2^{\circ}C$ on Wednesday

Net drop in temperature $1-(-2)=3^{\circ}C$

The temperature drops twice from Wednesday to Thursday

$\Delta T=2\times 3=6^{\circ}C$

Temperature on Thursday

$\Rightarrow -2-6=-8^{\circ}C$

3 0

3 years ago