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3 years ago

What is 1/6+1/2 in its simplest form?

Mathematics

2 answers:

Salsk061 [2.6K]3 years ago

6 0

1/2. 1/6
1/2×3=3/6
3/6+1/6=4/6
4/6=2/3

2/3

myrzilka [38]3 years ago

6 0

You have to find the common denominator for the 2 fractions.
leave 1/6 the same and turn ½ into 3/6.
then you can add 1/6+3/6 and get 4/6
then divide the numerator and the denominator by 2 any get ⅔
so ⅔ is your answer
Hope I helped :)

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What is the minimum number of clients the travel agent should survey? Note that z=1.96 for a 95% confidence interval.

slega [8]

Answer:

121

Step-by-step explanation:

Given data as per the question

Standard deviation = $\sigma$ = 840

Margin of error = E = 150

Confidence level = c = 95%

For 95% confidence, z = 1.96

based on the above information, the minimum number of clients surveyed by the travel agent is

$n = (\frac{z\times \sigma}{E})^2$

$= (\frac{1.96\times 840}{150})^2$

= 120.47

= 121

hence, the 121 number of clients to be surveyed

Therefore we applied the above formula to determine the minimum number of clients

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3 years ago

Step by step on how to solve this equation 2 x + 3 = x − 4 because I fogot how to do basic math

valentina_108 [34]

Answer:

-7

Step-by-step explanation:

2x +3 =x-4

2x - x =-4 - 7

x=-7

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3 years ago

A juice factory offers its juice in both bottles and cans. A class of students on a field trip decides to compare the numbers of

Verizon [17]

Answer: 50 cases per minute

Step-by-step explanation:

1. Select one of the points on the graph. Let's use (2, 100) in this case.

2. Divide the number of cases (y) by the amount of minutes (x) This means the equation would be 100 ÷ 2.

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3 years ago

Read 2 more answers

. (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.

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Elden [556K]

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