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Fudgin [204]
3 years ago
13

Wendell paid $12.85 for sunglasses and $2.75 for chips and a drink. How much did Wendell spend?

Mathematics
2 answers:
GrogVix [38]3 years ago
7 0
$15.60

Hope this helped

erma4kov [3.2K]3 years ago
4 0
15.6 dollars s how much money Wendell spent

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PLEASE HELP!!! At the state fair 6 employees were paid $6.93 an hour. They worked for 9 hours at regular price and then for 5 ho
castortr0y [4]

Answer:

Your answer should be 443.82

Step-by-step explanation:

Ok, so if 6 workers were paid 6.93 an hour each and worked 9 hours each then you would do 6.93 x 9 for the number of hours and the pay for each hour, which is 62. 37. Then you would times 62.37 by 6 because you have 6 workers and the question asks for how much they made in TOTAL. 62.37 x 6 is 344.22, then you still have you overtime hours. So if each worker makes 3.32 for every overtime hour then you would do 3.32 x 5 for the number of hours. And that equals 16.60, then you would multiply that by 6 for the number of workers, 16.60 x 6 = 99.60. So in total for regual hours they made 344.22, and in total for overtime hours they made 99.60. Now you add those to to get you overall  price which is 443.82. And that should be your answer

Hope I did some good :) Have a great day!

7 0
3 years ago
Volume of a cube is 343 cm cube . find the edge​
Charra [1.4K]

Answer:

d

Step-by-step explanation:

d

6 0
3 years ago
Read 2 more answers
Rico ran a race that was 3 km long. how many meters did rico run?
pishuonlain [190]
3000 meters in total.
4 0
3 years ago
The mayor is interested in finding a 98% confidence interval for the mean number of pounds of trash per person per week that is
saul85 [17]

Solution :

Given :

Sample mean, $\overline X = 34.2$

Sample size, n = 129

Sample standard deviation, s = 8.2

a. Since the population standard deviation is unknown, therefore, we use the t-distribution.

b. Now for 95% confidence level,

   α = 0.05, α/2 = 0.025

  From the t tables, T.INV.2T(α, degree of freedom), we find the t value as

  t =T.INV.2T(0.05, 128) = 2.34

  Taking the positive value of t, we get  

  Confidence interval is ,

  $\overline X \pm t \times \frac{s}{\sqrt n}$

 $34.2 \pm 2.34 \times \frac{8.2}{\sqrt {129}}$

 (32.52, 35.8)

 95% confidence interval is  (32.52, 35.8)

So with $95 \%$ confidence of the population of the mean number of the pounds per person per week is between 32.52 pounds and 35.8 pounds.

c. About $95 \%$ of confidence intervals which contains the true population of mean number of the pounds of the trash that is generated per person per week and about $5 \%$ that doe not contain the true population of mean number of the pounds of trashes generated by per person per week.  

4 0
3 years ago
I need help please thanks
maksim [4K]

Answer:

Equation: \frac{4}{3}x - 3

Step-by-step explanation:

Slope = 4/3

Y-intercept = -3

Therefore;

Equation: y = mx + b

Equation: \frac{4}{3}x - 3

Hope this helps!

6 0
3 years ago
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