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3 years ago

WHOEVER ANSWERS CORRECTLY I WILL MARK BRAINLIEST!!

Mathematics

2 answers:

Tom [10]3 years ago

7 0

Answer:

Area=27.58 cm^2

Perimeter =22.29cm

Step-by-step explanation:

The shape consists of a sector and a trapezium

For area= area of sector +area of trapezium

[θ/360*πr^2]+[1/2(a+b)h]

[40/360*3.14*4^2]+[1/2(4+7)4]

5.58+22

27.58cm^2

For perimeter = perimeter of sector +perimeter of square +perimeter of triangle

[θ/360*2πr]+[4s]+[1/2(a+b)]

[40/360*2*3.14*4]+[4*4]+[1/2(4+3)]

2.79+16+3.5

22.29cm

gizmo_the_mogwai [7]3 years ago

3 0

Answer:

Area = 342 $cm^2$ , Perimeter = 22.79 cm

Step-by-step explanation:

First, divide the shape into 3:

An arc, a square, and a triangle.

Dimensions of shapes:

Arc:

Degree = 40°

Arms = 4 cm

Square:

All sides are 4 cm

Triangle:

base = 3 cm

height = 4 cm

hypotenuse = 5 cm

To get perimeter:

Arc length = $\frac{40}{360} * (2*\pi * (4))$ (4 is the arm length ) = 2.7925 cm

Arc arm = 4 cm

Square sides (only two are outside) = 2 * 4 = 8 cm

Triangle (only hypotenuse and base are outside) = 5 + 3 = 8 cm

So total perimeter is 2.7925 + 4 + 8 + 8 = 22.79cm

To get area:

Arc area (the pizza shape) = $\frac{40}{2\pi } * \pi * (4)^{2}$ (4 is the arm length) = 320 $cm^{2}$

Square area = side * side = 4 * 4 = 16 $cm^{2}$

Triangle = (1/2) * base* height = (1/2) * 3 * 4 = 6 $cm^{2}$

So total area is 320 + 16 + 6 = 342 $cm^2$

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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 $t_{\alpha}= -2.571$

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Answer:

Step-by-step explanation:

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Combine like terms

5x2+4x-5 which is C

You are welcome and don't for get to click <u>THANKS</u>

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