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stich3 [128]
2 years ago
15

Plz help I need help on this problem

Mathematics
2 answers:
Evgen [1.6K]2 years ago
8 0

Answer:

y = 1/2(x) + 2

Step-by-step explanation:

y- 2 = 1/2(x)

y = 1/2(x) + 2

Answer:

y = 1/2(x) + 2

Elodia [21]2 years ago
6 0

Answer:

what is the problem

Step-by-step explanation:

i need picture

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Vika [28.1K]

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noseeee

Step-by-step explanation:

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3 years ago
X - y = -2<br> y = -x - 2<br> What is the solution
LenKa [72]

Answer:

<em>x </em>= -2

<em>y </em>= 0

Step-by-step explanation:

Pretty hard process, but let me explain:

First, solve for y.

Substitute -x - 2 for y in x - y = -2:

Then, simplify both sides of the equation.

x - y = -2

x - (-x - 2) = -2

After, add -2 to both sides of the equation.

2x + 2 = -2

2x + 2 + -2 = -2 + -2

Next, divided both sides by 2.

2x/2 = -4/2

After that, substitute -2 for x in y = -x - 2:

You must simplify both sides of the equation.

y = -x - 2

y = -(-2) - 2

y = 0

Sorry if this is long and confusing, but I hope this helps!!

4 0
3 years ago
What is 3 to the powers of 5
My name is Ann [436]

Answer:

3 to the power of 5 = 35 = 243

Step-by-step explanation:

4 0
3 years ago
What is the equvilent fraction of 7•4 OVER 1002•9
galina1969 [7]
14 over 4509 would be an equvilent fraction
7 0
2 years ago
Read 2 more answers
A random sample of 16 one-kilogram sugar packets is obtained and the actual weights of the packets are measured. The sample mean
elena55 [62]

Answer:

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

Step-by-step explanation:

We are in posession of the sample's standard deviation, so we use the student's t-distribution to find the confidence interval.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 16 - 1 = 15

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 35 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.99}{2} = 0.905([tex]t_{995}). So we have T = 2.9467

The margin of error is:

M = T*s = 2.9467*0.058 = 0.171

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 1.053 - 0.171 = 0.882kg

The upper end of the interval is the sample mean added to M. So it is 1.053 + 0.171 = 1.224 kg.

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

3 0
3 years ago
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