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podryga [215]
3 years ago
6

Evaluate 3{5+3[10+4•8]}​

Mathematics
1 answer:
Hatshy [7]3 years ago
8 0
The answer to your equation is 1008
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543x​=244x​+255x​
RUDIKE [14]

Answer:

The answer is x = 0

8 0
3 years ago
11x + 6y = 6 then graph the equation ​
maw [93]

Answer:

Step-by-step explanation:

Use the intercept method of graphing a straight line:

Let x = 0.  We get y = 1.  This is the y-intercept (0, 1).

Let y = 0.  We get x = 6/11.  This is the x-intercept (6/11, 0).

Plot both points and then draw a straight line through them.

6 0
3 years ago
Solve -11 - x + = - 5 (2x-3) + 7
garri49 [273]

Answer:

Let's see what to do buddy...

Step-by-step explanation:

_________________________________

- 11 - x =  - 5(2x - 3) + 7

- 11 - x =  - 10x + 15 + 7

- 11 - x =  - 10x + 22

Subtract the sides of the equation plus<em> </em><em>1</em><em>1</em> :

- x =  - 10x + 22 + 11

- x =  - 10x + 33

Subtract the sides of the equation plus <em>1</em><em>0</em><em>x</em>

10x - x = 10x - 10x + 33

9x = 33

Divided the sides of the equation by <em>9</em><em> </em>

\frac{9}{9}x =  \frac{33}{9} \\

x =  \frac{33}{9} =  \frac{3 \times 11}{3 \times 3} =  \frac{11}{3} \\

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

7 0
2 years ago
Criterion: Identify IV, DV, and hypotheses and evaluate the null hypothesis for an independent samples t test. Data: Use the inf
Papessa [141]

The table is missing in the question. The table is provided here :

Group 1        Group 2

 34.86            64.14      mean

 21.99            20.46      standard deviation

  7                    7                n

Solution :

a). The IV or independent variable = Group 1

    The DV or the dependent variable = Group 2

b).

  $H_0: \mu_1 = \mu_2$

  $H_a:\mu_1 < \mu_2$

  Therefore,   $t = \frac{\bar x_1 - \bar x_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{S_2^2}{n_2}}}$

$t = \frac{34.86 - 64.14}{\sqrt{\frac{21.99^2}{7}+\frac{20.46^2}{7}}}$

t = -2.579143

Now,   $df = min(n_1 - 1, n_2 - 1)$

           df = 7 - 1

               = 6

Therefore the value of p :

  $=T.DIST(-2.579143,6,TRUE)$

 = 0.020908803

The p value is 0.0209

$p< 0.05$

So we reject the null hypothesis and conclude that $\mu_1 < \mu_2$

7 0
3 years ago
7) what is the BEST conclusion we can draw from the graph <br><br> (the answers are in the picture)
Viktor [21]

Answer:

a

Step-by-step explanation:

7 0
3 years ago
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