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rewona [7]
3 years ago
12

PLEASE HELP IVE BEEN WAITING FOR 3 HOURS AND NO ANSWER

Mathematics
2 answers:
aleksandrvk [35]3 years ago
5 0
This situation is governed by a linear equation:

Total Pay = Base Salary + Commissions, and here that equation is:

Total Pay = $150 + 0.14(Total of sales for the week).

Here, Total Pay = $150 + 0.14($6050) = $150 + $847 = $997
REY [17]3 years ago
3 0
First, you need to create an equation for this problem. Remember 14% = 0.14
($6,050 × 0.14 + $6,050) + $150

Now, all you would have to do is solve it. You would start in the parenthesis using the order of operations (PEMDAS). 
($6,050 × 0.14 + $6,050) + $150
$6,897 + $150 = $<span>7,047
</span>
His total pay last week was $7,047. 

I hope this helps!
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A credit card company offers a cash back program on transactions. For every transaction over​ $100 the customer gets ​$3 back. F
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5 0
3 years ago
Read 2 more answers
How many photons with 10 ev are required to produce 20 joules of energy?
alexira [117]
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7 0
3 years ago
NEED ANSWERS ASAP, 100 POINTS! There are twenty classes at each of two middle schools. The number of students in each class at e
DiKsa [7]

Note: You missed to add the dot plots chart. As I found the chart after a little research. Thus, I am attaching it and based on that dot plot chart I am solving the question which anyways would clear you concept.

Answer:

'There are about 2 more students in each class at Oak Middle School than at Poplar Middle School' is the correct statement.

Step-by-step explanation:

From the diagram, it is clear that

The data set containing Poplar Middle School:

20   20   20   21   21   21   21   21   22   22   22   22   22   22   22   23   23   23   23   24  

The mean of a data set is the sum of the terms divided by the total number of terms. Using math notation we have:

\mu = 435/20

  =\frac{87}{4}

   = 21.75

The data set containing Oak Middle School:

20   21   21   22   22   23   23   23   23   24   24   24   25   25   26   26   27   27   28   29  

The mean of a data set is the sum of the terms divided by the total number of terms. Using math notation we have:

\mu = 483/20

   =24.15

So, the difference in mean will be:

                                            24.15 - 21.75 = 2.4

Therefore, 'there are about 2 more students in each class at Oak Middle School than at Poplar Middle School' is the correct statement.

5 0
3 years ago
A fair die is cast four times. Calculate
svetlana [45]

Step-by-step explanation:

<h2><em><u>You can solve this using the binomial probability formula.</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows:</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: </u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) </u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008Add them up, and you should get 0.1319 or 13.2% (rounded to the nearest tenth)</u></em></h2>
8 0
3 years ago
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