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3 years ago

6

7th grade math help me plzzz

1 answer:

Kaylis [27]3 years ago

7 0

Answer:

Step-by-step explanation:

I hope this helps

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Arjay, Dorothy, Melissa, and Gray live in the same city. Arjay and Dorothy live 2 miles from each other. Dorothy and Melissa liv

GarryVolchara [31]

Dorothy and Gray live 8 miles from each other.

5 0

3 years ago

Over What Interval is The Function In This Graph Increasing?

irina1246 [14]

The correct answer is option D.which is -4 ≤ x ≤ 1.

<h3>What is a graph?</h3>

A graph is the representation of the data on the vertical and horizontal coordinates so we can see the trend of the data.

in the graph, we can see that the value of x is increasing from -4 coordinate of x to the 1 coordinate of x so the interval will be given as -4 ≤ x ≤ 1.

Therefore the correct answer is option D.which is -4 ≤ x ≤ 1.

To know more about graphs follow

brainly.com/question/25020119

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1 year ago

Use the given similar triangles to find the value of x<br> I need the answer fast please!!

Yuki888 [10]

Answer:

I believe the answer is x=7

Step-by-step explanation:

<em>3x</em> and<em> 4x-1 </em> are corresponding sides and so are <em>14</em> and <em>18</em>.

You cross multiply with the smaller triangle lengths at the top.

3x/(4x-1)= 14/18

Cross multiply to get

18(3x)= 14(4x-1)

54x=56x-14

Subtract 54x on the other side and add 14 to the other (or subtract 56x to the other side)

you get 14=2x and divide 14 by 2 to get x=7

8 0

3 years ago

What's equivalent to y^9 - y^3

Yuki888 [10]

Y^9-y^3 can also be written as y^9-y^3=x
so it is also equivalent to x+y^3=y^9
and x-y^9=-y^3

7 0

3 years ago

Two shooters, Rodney and Philip, practice at a shooting range. They fire rounds each at separate targets. The targets are mark

Sedaia [141]

Complete Question

Answer:

a

$SE = 0.66}$

b

$-3.29 < \mu_1 - \mu_2 < -0.70$

Step-by-step explanation:

From the question we are told that

The sample size is n = 60

The first sample mean is $\= x _1 = 8$

The second sample mean is $\= x _2 = 10$

The first variance is $v_1 = 0.25$

The first variance is $v_2 = 0.55$

Given that the confidence level is 95% then the level of significance is 5% = 0.05

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the first standard deviation is

$\sigma_1 = \sqrt{v_1}$

=> $\sigma_1 = \sqrt{0.25}$

=> $\sigma_1 = 0.5$

Generally the second standard deviation is

$\sigma_2 = \sqrt{v_2}$

=> $\sigma_2 = \sqrt{0.55}$

=> $\sigma_2 = 0.742$

Generally the first standard error is

$SE_1 = \frac{\sigma_1}{\sqrt{n} }$

$SE_1 = \frac{0.5}{\sqrt{60} }$

$SE_1 = 0.06$

Generally the second standard error is

$SE_2 = \frac{\sigma_2}{\sqrt{n} }$

$SE_2 = \frac{0.742}{\sqrt{60} }$

$SE_2 = 0.09$

Generally the standard error of the difference between their mean scores is mathematically represented as

$SE = \sqrt{SE_1^2 + SE_2^2 }$

=> $SE = \sqrt{0.06^2 +0.09^2 }$

=> $SE = 0.66}$

Generally 95% confidence interval is mathematically represented as

$(\= x_1 -\= x_2) -(Z_{\frac{\alpha }{2} } * SE) < \mu_1 - \mu_2 < (\= x_1 -\= x_2) +(Z_{\frac{\alpha }{2} } * SE)$

=> $(8 -10) -(1.96 * 0.66) < \mu_1 - \mu_2 < (8-10) +(Z_{\frac{\alpha }{2} } * 0.66)$

=> $-3.29 < \mu_1 - \mu_2 < -0.70$

5 0

3 years ago