3. Below is a scholar's work to determine if T = 10 and W = 1575 is a solution

TRYING TO GRADUATE!

FrozenT [24]

There are 6 total possible outcomes. Rolling a 1, 2, or 3 is 3/6 or 1/2

8 0

3 years ago

I really need help ASAP Answer choices: 101,19,50,60,79 A= B= C= D= E=

FromTheMoon [43]

Answer:

Step-by-step explanation:

1) The sum of the angles on a straight line is 180 degrees. This means that

51 + b + 110 = 180

161 + b = 180

b = 180 - 161 = 19 degrees

2) Angle a = 60 degrees. This is so because they are vertically opposite angles.

3)The sum of angles in a triangle is 180 degrees. Therefore

angle a + angle b + angle c

60 + 19 + c = 180

79 + c = 180

c = 180 - 79 = 101 degrees

4) d + c = 180(sum of the angles on a straight line is 180 degrees). Therefore

d + 101 = 180

d = 180 - 101 = 79 degrees.

5) e + 51 + b + a = 180( sum of the angles in a triangle is 180 degrees). Therefore

e + 51 + 60 + 19 = 180

e + 130 = 180

e = 180 - 130 = 50 degrees

6 0

4 years ago

If s = 2 feet, then what is the surface area of the cube shown above? A. 24 square feet B. 12 square feet C. 16 square feet D. 4

lakkis [162]

Answer:

I think it is B

Step-by-step explanation:

I hope this helps.

7 0

3 years ago

Suppose that Y1, Y2,..., Yn denote a random sample of size n from a Poisson distribution with mean λ. Consider λˆ 1 = (Y1 + Y2)/

Burka [1]

Answer:

The answer is " $\bold{\frac{2}{n}}$ ".

Step-by-step explanation:

considering $Y_1, Y_2,........, Y_n$ signify a random Poisson distribution of the sample size of n which means is λ.

$E(Y_i)= \lambda \ \ \ \ \ and \ \ \ \ \ Var(Y_i)= \lambda$

Let assume that,

$\hat \lambda_i = \frac{Y_1+Y_2}{2}$

multiply the above value by Var on both sides:

$Var (\hat \lambda_1 )= Var(\frac{Y_1+Y_2}{2} )$

$=\frac{1}{4}(Var (Y_1)+Var (Y_2))\\\\=\frac{1}{4}(\lambda+\lambda)\\\\=\frac{1}{4}( 2\lambda)\\\\=\frac{\lambda}{2}\\$

now consider $\hat \lambda_2$ = $\bar Y$

$Var (\hat \lambda_2 )= Var(\bar Y )$

$=Var \{ \frac{\sum Y_i}{n}\}$

$=\frac{1}{n^2}\{\sum_{i}^{}Var(Y_i)\}\\\\=\frac{1}{n^2}\{ n \lambda \}\\\\=\frac{\lambda }{n}\\$

For calculating the efficiency divides the $\hat \lambda_1 \ \ \ and \ \ \ \hat \lambda_2$ value:

Formula:

$\bold{Efficiency = \frac{Var(\lambda_2)}{Var(\lambda_1)}}$

$=\frac{\frac{\lambda}{n}}{\frac{\lambda}{2}}\\\\= \frac{\lambda}{n} \times \frac {2} {\lambda}\\\\ \boxed{= \frac{2}{n}}$

8 0

3 years ago

Please help me with this one

mylen [45]

9514 1404 393

Answer:

y < -2x +3
y > 1/2x -2

Step-by-step explanation:

To find the equation of each boundary line, look for the y-intercept. Then find the rise and run to the next grid intersection to the right that is crossed by that line. The equation for the line is ...

y = mx + b . . . . . . m = rise/run = slope; b = y-intercept.

__

The lines are both dashed, so the inequalities will not include the "or equal to" case. For line (1), the shading is below the line, so the appropriate inequality symbol is <. For line (2), the shading is above the line, so the appropriate inequality symbol is >.

__

Line 1

y-intercept = +3. m = -2/1 = -2

y < -2x +3

Line 2

y-intercept = -2. m = 1/2

y > 1/2x -2

3 0

3 years ago