Which statement is true for the equation 5n − 4 = 5n − 3?

. (0.5 point) We simulate the operations of a call center that opens from 8am to 6pm for 20 days. The daily average call waiting

SashulF [63]

Answer:

The 95% t-confidence interval for the difference in mean is approximately (-2.61, 1.16), therefore, there is not enough statistical evidence to show that there is a change in waiting time, therefore;

The change in the call waiting time is not statistically significant

Step-by-step explanation:

The given call waiting times are;

24.16, 20.17, 14.60, 19.79, 20.02, 14.60, 21.84, 21.45, 16.23, 19.60, 17.64, 16.53, 17.93, 22.81, 18.05, 16.36, 15.16, 19.24, 18.84, 20.77

19.81, 18.39, 24.34, 22.63, 20.20, 23.35, 16.21, 21.73, 17.18, 18.98, 19.35, 18.41, 20.57, 13.00, 17.25, 21.32, 23.29, 22.09, 12.88, 19.27

From the data we have;

The mean waiting time before the downsize, $\overline x_1$ = 18.7895

The mean waiting time before the downsize, s₁ = 2.705152

The sample size for the before the downsize, n₁ = 20

The mean waiting time after the downsize, $\overline x_2$ = 19.5125

The mean waiting time after the downsize, s₂ = 3.155945

The sample size for the after the downsize, n₂ = 20

The degrees of freedom, df = n₁ + n₂ - 2 = 20 + 20 - 2 = 38

df = 38

At 95% significance level, using a graphing calculator, we have; $t_{\alpha /2}$ = ±2.026192

The t-confidence interval is given as follows;

$\left (\bar{x}_{1}- \bar{x}_{2} \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}$

Therefore;

$\left (18.7895- 19.5152 \right )\pm 2.026192 \times \sqrt{\dfrac{2.705152^{2}}{20}+\dfrac{3.155945^2}{20}}$

(18.7895 - 19.5125) - 2.026192*(2.705152²/20 + 3.155945²/20)^(0.5)

The 95% CI = -2.6063 < μ₂ - μ₁ < 1.16025996668

By approximation, we have;

The 95% CI = -2.61 < μ₂ - μ₁ < 1.16

Given that the 95% confidence interval ranges from a positive to a negative value, we are 95% sure that the confidence interval includes '0', therefore, there is sufficient evidence that there is no difference between the two means, and the change in call waiting time is not statistically significant.

6 0

2 years ago

1. Solve the inequality.

Sauron [17]

Answer:

Your final answer is either

x≥-2 if your initial inequality was

6x+2≤2(5-x)

OR

x≤-2

if your initial inequality was

6x+2≥2(x-2)

Step-by-step explanation:

As shown you have an equality, not an inequality.

-6x+2=2(5-x) distribute through parenthesis

-6x+2=2(5)+2(-x)

-6x+2=10-2x add 2x to both sides

2x-6x+2=10-2x+2x

-4x+2=10 subtract 2 from both sides

-4x+2-2=10-2

-4x=8 divide both sides by -4

-4x/(-4) = 8/(-4)

x = -2

With the ≥ or ≤ sign you would solve the exact same way

except for the point where when dividing both sides by

-4 requires you to reverse the inequality.

Your final answer is either

x≥-2 if your initial inequality was

6x+2≤2(5-x)

OR

x≤-2

if your initial inequality was

6x+2≥2(x-2)

7 0

3 years ago

B. What are the second and third terms of this arithmetic sequence? 80, 페, 페, 125,...........

nataly862011 [7]

The 2nd and 3rd term of an AP is found to be (a₂ = 95) and (a₃ = 110).

<h3>What is the sequence of AP arithmetic progression?</h3>

In Arithmetic Progression, the difference between the two numerical orders is a fixed number (AP). Arithmetic Sequence is another name for it.

We'd come across a few key concepts in AP that had been labeled as:

The first term (a)
Common difference (d)
Term nth (an)
The total of first n terms (Sn)

As shown below, the AP can also be referred to in terms of common differences.

The following is the procedure for evaluating an AP's n-th term: an = a + (n − 1) × d
The arithmetic progression sum is as follows: Sn = n/2[2a + (n − 1) × d].
Common difference 'd' of an AP: d = a2 - a1 = a3 - a2 = a4 - a3 = ...... = an - an-1.

Now, the given sequence is; 80, _, _, 125.

The series comprises of four given terms.

Let the first term be 'a₁' = 180.

The second term be 'a₂'.

The third term be 'a₃'.

And, the fourth term is 'a₄' = 125.

Use the nth term formula to find the common difference 'd'.

n-th term: an = a + (n − 1) × d

a₄ = a + (n - 1)d

125 = 80 + (4 - 1)d

45 = 3d

d = 15

Thus, the common difference is 15.

The second term is calculated as;

a₂ = a₁ + d

a₂ = 80 + 15

a₂ = 95.

The third term is estimated as;

a₃ = a₂ + d

a₃ = 95 = 15

a₃ = 110

Therefore, the 2nd and third term of an AP is computed as 95 and 110.

To know more about Arithmetic Sequence, here

brainly.com/question/24989563

#SPJ4

8 0

1 year ago

I have no idea how to answer this question I've tried everything look at the Picture

zlopas [31]

<u>Answer:</u>

a. the error is that he wrote 0.05 instead of 0.2

0.148835 ≈ 0.2

b.

0.2 = 2 × 0.1

0.2 = 2 × 10⁻¹

The mass is about 2 × 10⁻¹ kg

8 0

3 years ago

Jupiter's orbit is about 3×10^9 miles and Earth's orbit is about 6×10^8 miles. How many times longer is Jupiter's orbit than Ear

Tasya [4]

Answer:

5

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers