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Arturiano [62]
2 years ago
8

Answer gets brainliest

Mathematics
1 answer:
vlabodo [156]2 years ago
6 0

x= 3/2 and y=4

Hope this helps

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QUICK !!<br> What is the image point of (-9,5) after a translation right 3 units and down 2 units ?
mario62 [17]

Answer:

(-6, 3)

Step-by-step explanation:

What you do is based off the image point (-9, 5). What you do is you go 3 units right which is like -9+3=-6. Then you will go 2 units down which is like 5-2=3.

(-6, 3) is your answer.

4 0
3 years ago
Write the inequality for the linear inequality graphed below. You can
romanna [79]

Answer:

 

Step-by-step explanation:

5 0
2 years ago
Using f(x) = 2x + 7 and g(x) = x - 3, find f(g(x)).
Eva8 [605]

Answer:

2x+1

Step-by-step explanation:

that's a hard question

we know that g(x)= x-3

so f(g(x))= f(x-3)

we put it in the equation :

f(x-3)= 2(x-3) +7 = 2x-6+7 = 2x +1

6 0
3 years ago
How do i graph q is greater than -1.3 on a number line
kompoz [17]

Answer:

an open circle on -1.3 on the line and draw a line to the right with an arrow.

Step-by-step explanation:

On your number line, draw an open circle on -1.3 and then draw a line on the number line going to the right and ending with an arrow at the end.  The open circle means that it is that starting point, but is not equal to it.

5 0
3 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
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