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aleksandrvk [35]
3 years ago
8

What is the equation of the line that passes through the point (5, -2) and has a slope of 6/5?

Mathematics
1 answer:
kozerog [31]3 years ago
6 0

Answer:

y = 6/5x + 8

Step-by-step explanation:

m = 6/5

y - y1 = m (x - x1)

y - (-2) = 6/5(x - 5)

y + 2 = 6/5 x - 6

y = 6/5x + 8

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Leo had $91, which is 7 times as much money as Alison had. How much
NARA [144]

Answer:

7x=91

$13.

Step-by-step explanation:

Let x represent money that Alison has.

We have been given that Leo has 7 times as much money as Alison had. So the amount of money that Leo has, would be 7x.

We are also told that Leo had $91, so we will equate 7x with 91 and solve for x as:

7x=91

To solve for x, we will divide both sides by 7:

\frac{7x}{7}=\frac{91}{7}

x=13

Therefore, Alison has $13.

8 0
3 years ago
Read 2 more answers
10 points help fast please
Anna71 [15]

Answer: the slope will become negative

Step-by-step explanation: the slope becomes negative due to the negative sign and the slope line will go down left to right

6 0
3 years ago
Choose the line that is the best fit for the data.
Svet_ta [14]

Answer:

I would say line 2

Step-by-step explanation:

just think it is

3 0
2 years ago
The contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces. We are interested in testing to d
NikAS [45]

Answer:

Option b. should not be rejected

Step-by-step explanation:

We are given that the contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces.

We have to test whether the variance of the population is significantly more than 0.003, i.e.;

  Null Hypothesis, H_0 : \sigma = \sqrt{0.003}

Alternate Hypothesis, H_1 : \sigma > \sqrt{0.003}

The test statistics used here for testing variance is;

          T.S. = \frac{(n-1)s^{2} }{\sigma^{2} } ~ \chi^{2}__n_-_1

where, s = sample standard deviation = 0.06

           n = sample size = 26 cans

So, Test statistics = \frac{(26-1)0.06^{2} }{0.003 } ~ \chi^{2}__2_5

                            = 30

So, at 5% level of significance chi square table gives critical value of 37.65 at 25 degree of freedom. Since our test statistics is less than the critical so we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that null hypothesis should not be rejected and variance of population is 0.003.

5 0
3 years ago
How to solve this problem
PolarNik [594]
X- 6/4=7
x-6=7×4
x-6=28
x=28-6
x=22
4 0
3 years ago
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