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2 years ago

Find the slope and y-intercept of the graph of the equation. y= 2x + 7

Mathematics

1 answer:

faust18 [17]2 years ago

5 0

Slope 2
y- intercept -7

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Engineers want to design passenger seats in commercial aircraft so that they are wide enough to fit 95 percent of adult men. Ass

zimovet [89]

Answer:

$z=1.64$

And if we solve for a we got

$a=14.4 +1.64*1.1=16.204$

The 95th percentile of the hip breadth of adult men is 16.2 inches.

Step-by-step explanation:

Let X the random variable that represent the hips breadths of a population, and for this case we know the distribution for X is given by:

$X \sim N(14.4,1.1)$

Where $\mu=14.4$ and $\sigma=1.1$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.05$ (a)

$P(X$ (b)

We can find a quantile in the normal standard distribution who accumulates 0.95 of the area on the left and 0.05 of the area on the right it's z=1.64

Using this value we can set up the following equation:

$P(X$

$P(z$

And we have:

$z=1.64$

And if we solve for a we got

$a=14.4 +1.64*1.1=16.204$

The 95th percentile of the hip breadth of adult men is 16.2 inches.

8 0

2 years ago

The unit digit of a two digit number is one less than the ten digit number. If the number is increase by 8 and then divided by t

Colt1911 [192]

The tens digit is 3; the units digit is 3-1 = 2; and the number itself is 32.

<em>Lets simplify the problem,</em>

Let assume the "tens" digit be x

Then the "units" difference = (x-1), according to the condition.

Hence, the number itself is N = 10x + (x-1).

Then the number N+8 is 10x + (x-1) + 8 = 10x + x + 7 = 11x + 7.

From the last statement of the problem, we have this equation

$N+8 = 8*(x+y),or11x + 7 = 8*(x+(x-1)).$

Simplify and find "x"

$11x + 7 = 8*(2x-1)11x + 7 = 16x - 87 + 8 = 16x - 11x15 = 5xx = 15/5 = 3$ .

Thus the tens digit is 3; the units digit is 3-1 = 2; and the number itself is 32.

Learn more about UNIT Problems on:

brainly.com/question/12253743

#SPJ4

6 0

2 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

2 years ago

Factor: 9t^2-16u^2 ?

liq [111]

Answer:

Step-by-step explanation:

9t²-16u² = 3²t²-4²u² = (3t)²-(4u)² = (3t+4u)(3t-4u)

8 0

3 years ago

Go step by step to reduce the radical.<br> V216

oksano4ka [1.4K]

Answer:

$6\sqrt{6}$

Step-by-step explanation:

$\sqrt{216}$

factorise 216

$\sqrt{2*2*2*3*3*3}$

since it is square root (meaning it has index to) pair up the number

$2*3\sqrt{2*3$

$6\sqrt{6}$

6 0

3 years ago