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photoshop1234 [79]
2 years ago
5

Determine whether the system of equations has one solution, no solution, or infinitely many solutions. 4x=10+4y and 2x-2y=15

Mathematics
1 answer:
galben [10]2 years ago
6 0

Answer:

infinitely many solutions

Step-by-step explanation:

Given the simultaneous equation

4x=10+4y... 1

2x-2y=15 .... 2

_____________

4x-4y = 10 * 1

2x-2y = 15 * 2

__________--

4x-4y = 10

4x-4y = 30

Add both equation

8x - 8y = 10+30

8x-8y = 40

Divide though by 8

x-y = 5

x = 5 +  y

Since the result gave 1 equation and 2 unknowns, then we will let x =k

k = 5 + y

y = k -5

k can be any integer

(x,y) = (k, k-5)

This show that the equation has infinitely many solutions

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Answer:

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

The degrees of freedom are given by:

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Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

Step-by-step explanation:

Data given and notation  

\bar X=1.6 represent the sample mean

s=0.46 represent the sample deviation

n=32 sample size  

\mu_o =1.35 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:  

Null hypothesis:\mu =1.35  

Alternative hypothesis:\mu \neq 1.35  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

P-value  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

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Answer:

-2.5625

Step-by-step explanation:

<u>1)[-1/4/5] × [-7/10+0.95] - [-2/1/4] turn that into improper fractions</u>

2 1/4=9/4

which is 1 4/5(-7/10+0.95)9/4

<u>2) 1 4/5(-7/10+0.95)= 0.3125</u>

which is 0.3125-9/4

<u>3)Make into decimal form</u>

9/4 equals 2.25

which is 0.3125-2.25

which all equals -2.5625

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