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yulyashka [42]
2 years ago
9

Given that 2x+y= y+x+1 and y =2x+2y solve for x and y​

Mathematics
1 answer:
Pachacha [2.7K]2 years ago
7 0

Answer:

x=1,y=-2

Step-by-step explanation:

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Please help me :)<br> I need help
zzz [600]

Answer:

Option C

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7 0
2 years ago
Read 2 more answers
a movie theater sold a total of 148 students tickets and adult tickets. each student ticket cost 6$ and each adult tickets cost
LiRa [457]
Let A= adult tickets and S= student tickets
A+S=148
9A+6S=1,188
You can plug into a calculator or by hand is shown below
-6(A+S)=-6(148) Checking the answer
-6A-6S=-888 A=100
+ 100+S=148
9A+6S=1,188 S=48
——————— 9(100)+6(S)=1,188
3A+0S=300 900+6S=1,188
1/3(3A)=1/3(300) 6S=288
A=100 S=48

And to check is on the right side
3 0
2 years ago
A nutritionist has developed a diet that she claims will help people lose weight. Twelve people were randomly selected to try th
deff fn [24]

The diet which is developed by the nutritionist is effective at the 0.05 level of significance. Claim is agreed.

<h3>When do we use two-sample t-test?</h3>

The two-sample t-test is used to determine if two population means are equal.

A nutritionist has developed a diet that she claims will help people lose weight. In this,

  • Twelve people were randomly selected to try the diet.
  • Their weights were recorded prior to beginning the diet and again after 6 months.

Here are the original weights, in pounds, with the weight after 6 months in parentheses.

  • Before 192 212 171 215 180 207 165 168 190 184 200 196
  • After    183 196  174 211 160 191   162 175 190  179  189 195

The mean of the weights before 6 moths is,

\overline X_1=\dfrac{192+ 212 +171 +215 +180 +207 +165 +168 +190 +184 +200 +196 }{12}\\\overline X_1=190

The mean of the weights after 6 months is,

\overline X_2=\dfrac{ 183 +196  +174 +211 +160 +191   +162 +175 +190  +179  +189 +195  }{12}\\\overline X_1=183.75

Standard deviation of both the data is 16.9 and 14.7.

1. Null and Alternative Hypotheses.

The following null and alternative hypotheses need to be tested:

\begin{array}{ccl} H_0: \mu_1 & = & \mu_2 \\\\ \\\\ H_a: \mu_1 & > & \mu_2 \end{array}

This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

  • (2) Rejection Region

Based on the information provided, the significance level is α=0.05 and the degrees of freedom are df = 22. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

df_{Total} = df_1 + df_2 = 11 + 11 = 22

Hence, it is found that the critical value for this right-tailed test is

t_c=1.717, for α=0.05 and df=22

The rejection region for this right-tailed test is,

R = \{t: t > 1.717\}

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

t = \displaystyle \frac{\bar X_1 - \bar X_2}{\sqrt{ \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}(\frac{1}{n_1}+\frac{1}{n_2}) } }

\displaystyle \frac{ 190 - 183.75}{\sqrt{ \frac{(12-1)16.9^2 + (12-1)14.7^2}{ 12+12-2}(\frac{1}{ 12}+\frac{1}{ 12}) } } = 0.967

  • (4) Decision about the null hypothesis

Since it is observed that t=0.967≤tc=1.717 it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.1721 and since p=0.1721≥0.05p = 0.1721  it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis H₀ is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is greater than μ2​, at the α=0.05 significance level.

Confidence Interval

The 95% confidence interval is −7.16<μ<19.66

Thus, the diet which is developed by the nutritionist is effective at the 0.05 level of significance. Claim is agreed.

Learn more about the two-sample t-test here;

brainly.com/question/27198724

#SPJ1

8 0
2 years ago
How can I rename 82 thousands?
Setler [38]
82,000. hope that helped
6 0
3 years ago
A piggy bank contains 36 coins that consists of dimes and quarters totaling $5.85 how many quarters are in the piggy bank in you
Lesechka [4]
The main idea here is to "translate" the words into maths. 
First we need to identify the unknowns and label these. 

We need to know the number if dimes and the number of quarters. So lets say
x: number of dimes
y: number of quarters

Now lets write equations from the written problem. 
We know that there are 36 coind total, thus: 
x + y = 36
We also know that the coins total 5.85 dollars, but it is better to count in cents, that is 585 cents. 
x are the number of dimes, their value is x*10
y are quarters with value of y*25
thus: 
10x+25y=585

We have two equations and two unknowns now, that needs to be solved to get the answer. 

x + y = 36
10x+25y=585


4 0
3 years ago
Read 2 more answers
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