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zalisa [80]
3 years ago
7

6x + 7y = 1 8x – 3y = 25 Solve me fast explainly

Mathematics
1 answer:
Elenna [48]3 years ago
5 0

Answer:

{x,y} = {89/37,-71/37}

Step-by-step explanation:

 8x = 3y + 25

 [2]    x = 3y/8 + 25/8

Plug this in for variable  x  in equation [1]

  [1]    6•(3y/8+25/8) + 7y = 1

  [1]    37y/4 = -71/4

  [1]    37y = -71

Solve equation [1] for the variable  y  

  [1]    37y = - 71

  [1]    y = - 71/37

By now we know this much :

   x = 3y/8+25/8

   y = -71/37

Use the  y  value to solve for  x  

   x = (3/8)(-71/37)+25/8 = 89/37

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ruslelena [56]

Answer:

The answer is B 0,2

Step-by-step explanation:

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3 years ago
I’m so confused so can you help me
Lelu [443]
45²

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Try saying it aloud, it may help to hear it :)

This basically just means to multiply 45 by itself twice.

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6 0
4 years ago
Read 2 more answers
Assume that you have a sample of n 1 equals 6​, with the sample mean Upper X overbar 1 equals 50​, and a sample standard deviati
tigry1 [53]

Answer:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

df=6+5-2=9

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

n_1 =6 represent the sample size for group 1

n_2 =5 represent the sample size for group 2

\bar X_1 =50 represent the sample mean for the group 1

\bar X_2 =38 represent the sample mean for the group 2

s_1=7 represent the sample standard deviation for group 1

s_2=8 represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

We are assuming that the population variances for each group are the same

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic for this case is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

The pooled variance is:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

We can find the pooled variance:

S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67

And the pooled deviation is:

S_p=7.46

The statistic is given by:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

The degrees of freedom are given by:

df=6+5-2=9

The p value is given by:

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

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3 years ago
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Answer:

-5.5

Step-by-step explanation:

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Step-by-step explanation:

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