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3 years ago

Need help Asap

1 answer:

7 0

X = 44/23 and y= -5/23.

We want the coefficient of one of the variables to be the same in both equations; we can multiply the top equation by 3 and the bottom equation by 2 to make the coefficients of y equal:
3(10x-4y=20) → 30x-12y=60
2(8x+6y=14) → 16x+12y=28

This is possible due to the multiplication property of equality.

Now that the coefficients of y are the same, we can add the equations together (using the addition property of equality):
30x+16x = 60+28
46x = 88 (addition of like terms)

Divide both sides by 46:
46x/46 = 88/46 (division property of equality)
x = 88/46 = 44/23

Now we substitute x into the first equation:
10(44/23) - 4y = 20
440/23 - 4y = 20

It will be easier to solve this if we have a common denominator; 20 wholes = (20*23)/23 = 460/23;

440/23 - 4y = 460/23

Subtract 440/23 from both sides (subtraction property of equality):
440/23 - 4y - 440/23 = 460/23 - 440/23
-4y = 20/23

Divide both sides by -4 (division property of equality):
-4y/-4 = 20/23 ÷ -4/1
y = 20/23 × -1/4 = -20/92 = -5/23

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Step-by-step explanation:

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3 years ago

A group of eight individuals with high cholesterol levels were given a new drug that was designed to lower cholesterol levels. C

noname [10]

Answer:

Step-by-step explanation:

The data in the question is not well arranged. The correct arrangement is:

Before After difference

283 215

299 206

274 187

284 212

248 178

275 212

293 192

277 196

Solution:

a) This is a matched pair design/experiment. This is so because the measurement of cholesterol levels were carried out on the same individuals.

b) Corresponding cholesterol levels before and after treatment form matched pairs.

The data for the test are the differences between the cholesterol levels before and after treatment.

μd = cholesterol level before treatment minus cholesterol level after treatment.

Before After difference

283 215 68

299 206 93

274 187 87

284 212 72

248 178 70

275 212 63

293 192 101

277 196 81

Sample mean, xd

= (68 + 93 + 87 + 72 + 70 + 63 + 101 + 81)/8 = 79.4

xd = 79.4

Standard deviation = √(summation(x - mean)²/n

n = 8

Summation(x - mean)² = (68 - 79.4)^2 + (93 - 79.4)^2 + (87 - 79.4)^2 + (72 - 79.4)^2 + (70 - 79.4)^2 + (63 - 79.4)^2 + (101 - 79.4)^2 + (81 - 79.4)^2 = 1253.88

Standard deviation = √(1253.88/8)

sd = 12.52

For the null hypothesis

H0: μd ≥ 0

For the alternative hypothesis

H1: μd < 0

This is a one tailed test(left tailed test)

The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 8 - 1 = 7

The formula for determining the test statistic is

t = (xd - μd)/(sd/√n)

t = (79.4 - 0)/(12.52/√8)

t = 17.94

We would determine the probability value by using the t test calculator.

p < 0.00001

4) Assume alpha = 0.05

Since alpha, 0.05 > than the p value, then we would reject the null hypothesis. Therefore, at 5% level of significance, we cannot conclude that the mean cholesterol level after treatment is less than the mean before treatment.

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3 years ago

Which three-dimensional figure has the greatest number of faces?

Digiron [165]

Answer:

Octagonal pyramid

Step-by-step explanation:

Hexagonal Prism: 8 faces

2 hexagon bases, and 6 faces that connect each side of the bases

Octagonal pyramid: 9 faces

One octagon base, and 8 faces that meet at a point- all extending from each side of the base

Rectangular prism: 6 faces

Two rectangle bases, and 4 faces connecting each side of the bases

Pentagonal pyramid: 6 faces

One pentagon base, and 5 faces that meet at a point- all extending from each side of the base

I recommend searching for a picture of each of the figures to help you understand this.

3 0

3 years ago

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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

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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:

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

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

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.

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