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3 years ago

8

Find the value of the expression if x=-7

1 answer:

LUCKY_DIMON [66]3 years ago

5 0

Answer: -16

Step-by-step explanation:

49 would be the result of -7 squared. Due to the fact it is negative, it would cancel out to be positive.

For the numerator, 49-1 is equivalent to 48.

48/x+4

-7+4 = -3

48/-3

= -16

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A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex

quester [9]

Answer:1866

Step-by-step explanation:

Given

n=200

x=118

Population proportion P= $\frac{118}{200}$ =0.59

$\alpha$ =0.005

Realiability =99%

$Z_{\frac{\alpha }{2}}=2.576$

Margin of erroe is given by $\sqrt{\frac{p\left ( 1-p \right )}{N}}$

0.03= $\sqrt{\frac{0.59\left ( 1-0.59 \right )}{N}}$

85.667= $\sqrt{\frac{N}{0.6519}}[tex]N=1865.88[tex]\approx 1866 Students$

4 0

3 years ago

Find the mass of the lamina that occupies the region D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with the density function ρ(x, y) = xye

Alona [7]

Answer:

The mass of the lamina is 1

Step-by-step explanation:

Let $\rho(x,y)$ be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:

$m=\int\limits \int\limits_D \rho(x,y) \, dA$ .

From the question, the given density function is $\rho (x,y)=xye^{x+y}$ .

Again, the lamina occupies a rectangular region: D={(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

The mass of the lamina can be found by evaluating the double integral:

$I=\int\limits^1_0\int\limits^1_0xye^{x+y}dydx$ .

Since D is a rectangular region, we can apply Fubini's Theorem to get:

$I=\int\limits^1_0(\int\limits^1_0xye^{x+y}dy)dx$ .

Let the inner integral be: $I_0=\int\limits^1_0xye^{x+y}dy$ , then

$I=\int\limits^1_0(I_0)dx$ .

The inner integral is evaluated using integration by parts.

Let $u=xy$ , the partial derivative of u wrt y is

$\implies du=xdy$

and

$dv=\int\limits e^{x+y} dy$ , integrating wrt y, we obtain

$v=\int\limits e^{x+y}$

Recall the integration by parts formula: $\int\limits udv=uv- \int\limits vdu$

This implies that:

$\int\limits xye^{x+y}dy=xye^{x+y}-\int\limits e^{x+y}\cdot xdy$

$\int\limits xye^{x+y}dy=xye^{x+y}-xe^{x+y}$

$I_0=\int\limits^1_0 xye^{x+y}dy$

We substitute the limits of integration and evaluate to get:

$I_0=xe^x$

This implies that:

$I=\int\limits^1_0(xe^x)dx$ .

Or

$I=\int\limits^1_0xe^xdx$ .

We again apply integration by parts formula to get:

$\int\limits xe^xdx=e^x(x-1)$ .

$I=\int\limits^1_0xe^xdx=e^1(1-1)-e^0(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(-1)=1$ .

No unit is given, therefore the mass of the lamina is 1.

3 0

3 years ago

You have just used the network planning model and found the critical path length is 30 days and the variance of the critical pat

joja [24]

Answer:

0.726 is the probability that the project will be completed in 33 days or less.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 30 days

Variance = 25 days

Standard Deviation,

$\sigma = \sqrt{\text{Variance}} = \sqrt{25} = 5$

We assume that the distribution of path length is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(completed in 33 days or less)

$P( x \leq 33) = P( z \leq \displaystyle\frac{33 - 30}{5}) = P(z \leq 0.6)$

Calculation the value from standard normal z table, we have,

$P(x \leq 33) = 0.726 = 72.6\%$

0.726 is the probability that the project will be completed in 33 days or less.

8 0

3 years ago

Mr. Cruise drove 70 miles in March he drove seven times as many miles in March as he did in January he drove four times as many

SVEN [57.7K]

70/7= 10 miles in January
10/4 = 2.5 miles in February

5 0

3 years ago

WRITE a Linear FUNCTION, f, with the given values:<br> f(-4) = -5 and f(2) = -3

Crazy boy [7]

Answer:

answer to the question is A

8 0

3 years ago