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Olenka [21]
3 years ago
11

Alfredo is employed on an assembly line. He receives 35 cents for each part that he works on. If he works on more than 150 parts

, his employer will pay him 40 cents for each part over 150. Yesterday, Alfredo worked on 210 parts. How much did he earn?
SHOW ALL WORK
Mathematics
1 answer:
aleksandrvk [35]3 years ago
3 0

Answer:

$76.50

Step-by-step explanation:

Let's split this up.

First, we will do the 150 parts. But, we need to subtract so that we know how many parts he worked on over 150.

So...

210-150=60

Now that we know how many parts he worked on over 150, let's multiply.

.35*150=52.50

Now, let's multiply what he did over 150.

.40*60=24

Finally, we have to add his pay up.

52.50+24.00=76.50

Making your answer...

$76.50

<u><em>Hope this helps!!!</em></u>

<u><em>Brady</em></u>

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Step-by-step explanation:

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2 years ago
Select the correct answer. A company employs 48 people in various departments. The average annual salary of each employee is $25
Anna35 [415]

Answer:

A. $1,056,000 ≤ x ≤ $1,344,000

Step-by-step explanation:

The average annual salary of each of the 48 employees is given as;

$25,000

The total salary that the company pays to its employees annually is thus;

$25,000 * 48 = $1,200,000

Now, the total annual maximum variance  of the employees salaries would be;

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The lower limit of the total salary that the company pays to its employees annually is calculated as;

$1,200,000 - $144,000 = $1056000

The upper limit of the total salary that the company pays to its employees annually is calculated as;

$1,200,000 + $144,000 = $1344000

Therefore, the range of the total salary that the company pays to its employees annually is;

$1,056,000 ≤ x ≤ $1,344,000

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3 years ago
Work out the number of sides of a regular polygon with the interior angle of 172 degrees.
Harman [31]
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6. The mean number of hours of flying time for pilots at Continental Airlines is 37 hours per month (The Wall Street Journal, Fe
SashulF [63]

Answer:

The marginal error is 2.355

Confidence interval is (34.645,\ 39.355)

The LB and UB are 34.645 and 39.3555 respectively.

Step-by-step explanation:

Consider the provided information.

n = 49, s = 8.2 c = 95%and \bar x=37

Degree of freedom is: n-1 = 49-1 = 48.

\alpha =\frac{1-c}{2}= \frac{1-0.95}{2}\\\alpha = \frac{0.05}{2}= 0.025

From the table t_{\alpha /2}=2.010635

Marginal error is:

E=t_{\alpha /2}\times \frac{s}{\sqrt{n}}

E=2.0106\times \frac{8.2}{\sqrt{49}}

E=2.0106\times \frac{8.2}{7}

E=2.0106\times 1.171

E=2.355

Hence, the marginal error is 2.355.

Part (B)

95% confidence interval estimate

(\bar x-E,\ \bar x+E)

(37-2.355,\ 37+2.355)

(34.645,\ 39.355)

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Answer:

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If x^3 = 729, x = 9.

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