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lesantik [10]
2 years ago
8

Which equation can be used to represent "three minus the difference of a number and one equals one-half of the difference of thr

ee times the same number and four"?
(1 – n) – 3 = 6 plus 2 left-parenthesis x plus 4 right parenthesis equals StartFraction one-half EndFraction left-parenthesis 3 minus x right-parenthesis.(4 – 3n)

3 – (1 – n) = 6 plus 2 left-parenthesis x plus 4 right parenthesis equals StartFraction one-half EndFraction left-parenthesis x minus 3 right-parenthesis.(4 – 3n)

(n – 1) – 3 = left-parenthesis 6 plus 2 right-parenthesis left-parenthesis x plus 4 right parenthesis equals StartFraction one-half EndFraction left-parenthesis 3 minus x right-parenthesis.(3n – 4)

3 – (n – 1) = left-parenthesis 6 plus 2 right-parenthesis left-parenthesis x plus 4 right parenthesis equals StartFraction one-half EndFraction left-parenthesis x minus 3 right-parenthesis.(3n – 4)
Mathematics
2 answers:
Rina8888 [55]2 years ago
9 0

Answer:

3 - (n - 1) = 1/2(3n - 4)

Step-by-step explanation:

We want to write three minus the difference of a number and one equals one-half of the difference of three times the same number and four as an equation.

Let the number be n.

The first part is: three minus the difference of a number and one:

3 - (n - 1)

The second part is: one-half of the difference of three times the same number and four:

1/2(3n - 4)

Now, let us equate the first and second parts:

3 - (n - 1) = 1/2(3n - 4)

PS: I really do not understand the options

Ne4ueva [31]2 years ago
7 0

Answer:

D) 3 – (n – 1) = left-parenthesis 6 plus 2 right-parenthesis left-parenthesis x plus 4 right parenthesis equals StartFraction one-half EndFraction left-parenthesis x minus 3 right-parenthesis.(3n – 4)



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5) In a certain supermarket, a sample of 60 customers who used a self-service checkout lane averaged 5.2 minutes of checkout tim
Ne4ueva [31]

Answer:

\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643

S_p=2.940

t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751

df=60+72-2=130

p_v =P(t_{130}

Assuming a significance level of \alpha=0.05 we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

Step-by-step explanation:

Data given

Our notation on this case :

n_1 =60 represent the sample size for people who used a self service

n_2 =72 represent the sample size for people who used a cashier

\bar X_1 =5.2 represent the sample mean for people who used a self service

\bar X_2 =6.1 represent the sample mean people who used a cashier

s_1=3.1 represent the sample standard deviation for people who used a self service

s_2=2.8 represent the sample standard deviation for people who used a cashier

Assumptions

When we have two independent samples from two normal distributions with equal variances we are assuming that  

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

The statistic 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}}}

And t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

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

System of hypothesis

Null hypothesis: \mu_1 \geq \mu_2

Alternative hypothesis: \mu_1 < \mu_2

This system is equivalent to:

Null hypothesis: \mu_1 - \mu_2 \geq 0

Alternative hypothesis: \mu_1 -\mu_2 < 0

We can find the pooled variance:

\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643

And the deviation would be just the square root of the variance:

S_p=2.940

The statistic is given by:

t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751

The degrees of freedom are given by:

df=60+72-2=130

And now we can calculate the p value with:

p_v =P(t_{130}

Assuming a significance level of \alpha=0.05 we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

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