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3 years ago

Simplify. t^-6 * t^2

Mathematics

1 answer:

Deffense [45]3 years ago

8 0

<span>t^-6 * t^2
= t^-4

hope it helps</span>

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Solve the system 5x+3y=-9 & -2+y=8

Illusion [34]

X=-39/5 and y=10 :) i hope this helps you out !

5 0

4 years ago

Given f(x) and g(x) = f(x) + k, use the graph to determine the value of k.

zalisa [80]

<h3>Answer: C) 4</h3>

All points on the blue curve f(x) are shifted up 4 units to get corresponding points on the red curve g(x). For example, the point (0,-3) moves up 4 units to (0,1).

g(x) = f(x)+k

g(x) = f(x)+4

3 0

3 years ago

To test for the significance of a regression model involving 3 independent variables and 51 observations, the numerator and deno

LenKa [72]

Answer:

numerator degree of freedom = 3

Denominator degree of freedom = 47

Step-by-step explanation:

The numerator degree of freedom is given by :

p - 1 ; where p = number of predictors ;

p = number of independent variables + 1

Number of independent variables = 3

p = 3 + 1 = 4

Numerator degree of freedom = p - 1 = 4 - 1 = 3

The denominator degree of freedom = n - p ; where n = number of observations

Number of observations, n = 51

Denominator degree of freedom = n - p = 51 - 4 = 47

6 0

3 years ago

A teacher was interested in finding out whether a special study program would increase the scores of students on a national exam

bija089 [108]

Answer:

Check the explanation

Step-by-step explanation:

Going by the first attached image below we reject H_o against H_1 if obs. $T > t_{\alpha /2;n-1}$

here obs.T=1.879

$\therefore obs.T \ngtr 2.447=t_{0.025;6}$

we accept $H_o:\mu _{1}=\mu_{2}$ at 5% level of significance.

i.e there is no sufficient evidence to indicate that the special study program is more effective at 5% level of significance.

this problem is simillar to the previous one except the alternative hypothesis.

Let X_i's denote the bonuses given by female managers and Y_i's denote the bonuses given by male managers.

we assume that $X_i \sim N(\mu _{1},\sigma _{1}^{2}) Y_i \sim N(\mu _{2},\sigma _{2}^{2})$ independently

We want to test $H_0:\mu_{1}=\mu_{2} vs H_1:\mu_{1}\neq \mu_{2}$

define $D_i=X_i-Y_i , i=1(1)8$

now $D_i\sim N(\mu _{1}-\mu _{2}=\mu _{D},\sigma _{1}^{2}+\sigma _{2}^{2}=\sigma _{D}^{2}) , i=1(1)8$

the hypothesis becomes

$H_0:\mu_{D}=0 vs H_1:\mu_{D}\neq 0$

in the third attached image, we use the same test statistic as before

i.e at 5% level of significance there is not enough evidence to indicate a difference in average bonuses .

3 0

3 years ago

Evaluate the following expression 30x^2y x=-2/5 y=5/8

meriva

30(-2/5)²(5/8) = 30*(4/25)*(5/8) = 30*4/(5*8) = 3

8 0

3 years ago