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3 years ago

Evaluate the following expression: 3y2 - 4x + 2xy, when x = 2 and y = 5

Mathematics

1 answer:

Trava [24]3 years ago

8 0

Answer:

Step-by-step explanation:

3×(5)^2 - 4×2 + 2×2×5

3×(5)^2 - 4×2 + 2×2×53×25 - 8 + 20

3×(5)^2 - 4×2 + 2×2×53×25 - 8 + 2075 + 20 - 8 

3×(5)^2 - 4×2 + 2×2×53×25 - 8 + 2075 + 20 - 8 95-8

3×(5)^2 - 4×2 + 2×2×53×25 - 8 + 2075 + 20 - 8 95-887

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WILL GIVE BRAINIEST what is the sum of the interior angles of the polygon shown below.

jolli1 [7]

Answer:

900

Step-by-step explanation:

(n-2)*180

5*180=900

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3 years ago

Read 2 more answers

What is the area of an isosceles trapezoid, if its shorter base has length of 18 cm, the length of the altitude is 9 cm, and the

padilas [110]

The formula for the area of a triangle of base b and altitude h is A = (1/2)(b)(h).

Here, b = 18 cm and h = 9 cm. Thus, the desired area is:

(18 cm)(9 cm)

A = ----------------------- = 81 cm^2

This problem gives you more info than is needed to solve it. The area of the given triangle is 81 cm^2.

5 0

4 years ago

Use trig identities to transform the left side of the equation to the right side.

Taya2010 [7]

Answer and explanation

(1+sinθ)(1-sinθ)=cos²θ

We are to prove that the left hand side is equal to the right hand side.

(1+sinθ)(1-sinθ) = 1(1-sinθ) + sinθ(1-sinθ)

= 1 - sinθ + sinθ - sin²θ

= 1 - sin²θ

From the trigonometric identity sin²θ + cos²θ = 1,

1 - sin²θ = cos²θ

7 0

3 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 $\alpha =0.01$