Solve by the linear combination method. 2x + 3y = 11 –4x + 2y = 2 60pts please answer in 5 minutes

A 4 meter ladder leans against a house forming a

Ket [755]

Answer:

2.meters

Step-by-step explanation:

Lenght of ladder = 4 m

θ = 30°

To obtain h;

Using Pythagoras rule :

Sinθ = opposite / hypotenus

Hypotenus = 4 ; opposite = h

Sin 30 = h / 4

0.5 = h / 4

h = 0.5 * 4

h = 2 m

Hence, distance between th efoot of ladder and the base of the wall is 2 m

6 0

2 years ago

In a right triangle, the measure of one acute angle is 12 more than twice the measure of the other acute angle. Find the measure

soldi70 [24.7K]

Given:

In a right triangle, the measure of one acute angle is 12 more than twice the measure of the other acute angle.

To find:

The measures of the 2 acute angles of the triangle.

Solution:

Let x be the measure of one acute angle. Then the measure of another acute is (2x+12).

According to the angle sum property, the sum of all interior angles of a triangle is 180 degrees. So,

$x+(2x+12)+90=180$

$3x+102=180$

$3x=180-102$

$3x=78$

Divide both sides by 3.

$x=\dfrac{78}{3}$

$x=26$

The measure of one acute angle is 26 degrees. So, the measure of another acute angle is:

$2x+12=2(26)+12$

$2x+12=52+12$

$2x+12=64$

Therefore, the measures of two acute angles are 26° and 64° respectively.

8 0

2 years ago

How do we convert kilogram to gram

Sloan [31]

1kg=1000g

7kg*1000=7000g

10kg*1000=10000g

15g*1000=15000g

5 0

2 years ago

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases: a. Central area 5 .

Flauer [41]

Answer:

a) "=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got $t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228$

b) "=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got $t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086$

c) "=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got $t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845$

d) "=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got $t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678$

e) "=T.INV(1-0.01,25)"

And we got $t_{\alpha}= 2.485$

f) "=T.INV(0.025,5)"

And we got $t_{\alpha}= -2.571$

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

We will use excel in order to find the critical values for this case

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases:

a. Central area =.95, df = 10

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have $\alpha/2=0.025$ .

We can use the following excel codes:

"=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got $t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228$

b. Central area =.95, df = 20

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have $\alpha/2=0.025$ .

We can use the following excel codes:

"=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got $t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086$

c. Central area =.99, df = 20

For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have $\alpha/2=0.005$ .

We can use the following excel codes:

"=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got $t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845$

d. Central area =.99, df = 50

For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have $\alpha/2=0.005$ .

We can use the following excel codes:

"=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got $t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678$

e. Upper-tail area =.01, df = 25

For this case we need on the right tail 0.01 of the area and on the left tail we will have 1-0.01 = 0.99 , that means $\alpha =0.01$

We can use the following excel code:

"=T.INV(1-0.01,25)"

And we got $t_{\alpha}= 2.485$

f. Lower-tail area =.025, df = 5

For this case we need on the left tail 0.025 of the area and on the right tail we will have 1-0.025 = 0.975 , that means $\alpha =0.025$

We can use the following excel code:

"=T.INV(0.025,5)"

And we got $t_{\alpha}= -2.571$

8 0

3 years ago

Find the standard form of the equation of the parabola with a focus at (7, 0) and a directrix at x = -7.

Alecsey [184]

Answer:

y^2 = 28x

Step-by-step explanation:

Since the directrix is horizontal, use the equation of a parabola that opens left or right.

(y−k)^2 = 4p(x−h)

Find the vertex.

(0,0)

Find the distance from the focus to the vertex.

p = 7

Substitute in the known values for the variables into the equation

(y−k)^2 = 4p(x−h).

(y−0)^2 = 4(7)(x−0)

Simplify.

y^2 = 28x

7 0

3 years ago

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