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cluponka [151]
3 years ago
11

Which strategy should be used to correctly solve this problem?

Mathematics
1 answer:
hammer [34]3 years ago
6 0
B, make a table of possible values

the best way to do it is actually not listed
to have max area with minimu perimiter, try to get the sides as close measure to each other as possible


I would pick B though

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Mai biked 6\frac{3}{4} miles today, and Noah biked 4\frac{1}{2} miles.
Alinara [238K]

It was 1.5 times the length of Noah's bike ride as Mai’s bike ride.

The problem we dealing with is related to mixed numbers which are referred to as mixed numbers and could be the whole number, and a proper division spoke together. It for the most part speaks to a number between any two whole numbers.

Since we are providing mai's rate which is  6 3/4  miles and Noah's  rate of 4 1/2 miles, So in  order to find the time the length of Noah's ride is compared to Mal's, we just need to divide the given rates by each other

=> 6 3/4 divided by 4 1/2,  

=> 27/4 divided by 9/2, (The result is an improper fractions)

=> 27/4 X 2/9

=>3/2

=> 1.5

To know more about mixed numbers refer to the link  brainly.com/question/24137171?referrer=searchResults.

#SPJ1

3 0
1 year ago
Plz help me with my math
rodikova [14]

Answer: The answer is C

Step-by-step explanation:

8 0
3 years ago
A cereal company's cost, in thousands of dollars, is represented by the function C(x)=2x+4500 and its revenue, in thousands of d
jekas [21]

Given , revenue function : R(x)= 5x

Cost function C(x) = 2x+4500

Profit  = Revenue - Cost

Hence, P = R(x) - C(x)

Total revenue for x = 12000 is R(12000) = 5(12000) = 60000

Total cost for x = 12000, C(12000) = 2(12000)+4500 = 28500

Profit = 60000-28500 = 31500

Answer = 31500

6 0
3 years ago
How do u write 1.2 as a fraction
Ivahew [28]
Split the number into its whole number component and decimal component.

for the decimal component, 0.2 is the same as 2/10

Find the GCF of both 2 and 10 and that would be 2

divide both the numerator and denominator by the GCF 

simplify and you get 1/5

Lastly combine the whole number component with the fraction and your answer would be 1 1/5.
6 0
4 years ago
. (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
3 years ago
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