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monitta
3 years ago
10

Y=3x+2 i need a graph for this

Mathematics
1 answer:
klasskru [66]3 years ago
7 0

Graph is in the picture.

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Write a polynomial function that meets the given conditions. Answers may vary. Degree 2 polynomial with zeros of 2√ 13 and -2√13
MAVERICK [17]

Answer:

The polynomial function is x^{2} - 52

Step-by-step explanation:

A polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = (x - x_{1})*(x - x_{2}).

In this problem:

The roots are x_{1} = 2\sqrt{13} and x_{2} = -2\sqrt{13}

Then

(x - 2\sqrt{13}) \times (x - (-2\sqrt{13})) = (x - 2\sqrt{13}) \times (x + 2\sqrt{13}) = x^{2} - 2x\sqrt{13} + 2x\sqrt{13} -(2\sqrt{13})^{2} = x^{2} - 52

The polynomial function is x^{2} - 52

8 0
3 years ago
I didn’t do this but my friend needs help so help
yuradex [85]

Answer:

1. X = 14

2.X =13

Step-by-step explanation:

The angles are 90 degrees. I did math to check if the x values are correct and they are.

8 0
3 years ago
A random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6. A random sample of 17 su
Sladkaya [172]

Answer:

We conclude that there is no difference in potential mean sales per market in Region 1 and 2.

Step-by-step explanation:

We are given that a random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6.

A random sample of 17 supermarkets from Region 2 had a mean sales of 78.3 with a standard deviation of 8.5.

Let \mu_1 = mean sales per market in Region 1.

\mu_2  = mean sales per market in Region 2.

So, Null Hypothesis, H_0 : \mu_1-\mu_2 = 0      {means that there is no difference in potential mean sales per market in Region 1 and 2}

Alternate Hypothesis, H_A : > \mu_1-\mu_2\neq 0      {means that there is a difference in potential mean sales per market in Region 1 and 2}

The test statistics that will be used here is <u>Two-sample t-test statistics</u> because we don't know about population standard deviations;

                            T.S.  =  \frac{(\bar X_1 -\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+ {\frac{1}{n_2}}} }   ~  t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean sales in Region 1 = 84

\bar X_2 = sample mean sales in Region 2 = 78.3

s_1  = sample standard deviation of sales in Region 1 = 6.6

s_2  = sample standard deviation of sales in Region 2 = 8.5

n_1 = sample of supermarkets from Region 1 = 12

n_2 = sample of supermarkets from Region 2 = 17

Also, s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times  s_2^{2}  }{n_1+n_2-2} }  = s_p=\sqrt{\frac{(12-1)\times 6.6^{2}+(17-1)\times  8.5^{2}  }{12+17-2} } = 7.782

So, <u><em>the test statistics</em></u> =  \frac{(84-78.3)-(0)}{7.782 \times \sqrt{\frac{1}{12}+ {\frac{1}{17}}} }  ~   t_2_7

                                   =  1.943  

The value of t-test statistics is 1.943.

 

Now, at a 0.02 level of significance, the t table  gives a critical value of -2.472 and 2.473 at 27 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have<u><em> insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that there is no difference in potential mean sales per market in Region 1 and 2.

6 0
3 years ago
The authors of a paper presented a correlation analysis to investigate the relationship between maximal lactate level x and musc
ArbitrLikvidat [17]

Answer:

a) Sample correlation coefficient, r = 0.7411

bi) test statistic, t = 4.102

bii) P-value = 0.000736

Step-by-step explanation:

a) The formula for the sample correlation coefficient is given by the formula:

r = \frac{S_{xy} }{\sqrt{S_{xx} S_{yy} }} }

S_{xx} = 2,648,130.357\\S_{yy} = 36.7376,\\S_{xy} = 7408.225

r = \frac{7408.225}{\sqrt{2648130.357*36.7376} }

r = 0.7511

b)

i) formula for the test statistic is given by the formula:

t = \frac{r\sqrt{n-1} }{\sqrt{1 - r^{2} } }

sample size, n = 4

t = \frac{0.7511\sqrt{14-1} }{\sqrt{1 - 0.7511^{2} } }

t = 4.102

ii) Degree of freedom, df = n -2

df = 14 -2

df = 12

The P-value is calculate from the degree of freedom and the test statistic using excel

P-value =(=TDIST(t,df,tail))

P-value = (=TDIST(4.1,12,1)

P-value = 0.000736

4 0
3 years ago
A company wants to collect opinions on employee working conditions. Which group of individuals would be the best choice for a ra
makvit [3.9K]

Answer: B

Step-by-step explanation:

3 0
3 years ago
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