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3 years ago

If you were a salaried employee paid weekly, how many times would you be paid for an entire year?

Mathematics

2 answers:

Fudgin [204]3 years ago

6 0

Answer:

52 i think

Step-by-step explanation:

Natali [406]3 years ago

5 0

Answer: Most employers expect their exempt employees to work the number of hours necessary to get their jobs done. It doesn't matter if that takes more or fewer than 40 hours per week. Even if your exempt employee works 70 hours in a week, you are still only required to pay them their standard base salary If you use a payroll processing service, there is a cost for each check processed. But either way, processing paychecks more frequently will cost more, such as if you process paychecks every other week (26 times a year) rather than bi-monthly (24 times a year).

Step-by-step explanation:

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S is the midpoint of RT. If RS = 4x - 3 and RT = 6x - 4, what’s the value of ST?

Georgia [21]

Answer:

R__S__T

RT = RS + ST

6x – 4 = ( 4x – 3) + ST

(6x – 4 ) – ( 4x – 3 ) = ST

(6x – 4 ) +( – 4x + 3 ) = ST

2x – 1 = ST

ST = 2x – 1

RS=ST

4x – 3 = 2x – 1

4x – 2x = – 1 +3

2x = 2

x= 2/2

x =1

ST = 2x – 1 = 2(1) – 1 =2 ‐ 1 = 1

ST = 1

I hope I helped you^_^

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3 years ago

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love history [14]

Answer:

Function 1 has a greater slope and greater y-intercept than Function 2.

Step-by-step explanation:

Function 1

Slope: 1/3

y-intercept: 2

Function 2

Slope: -2/4

y-intercept: -2

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3 years ago

There are 11 executives and 90 employess. The executives makes $1000 per hour, and the employessw make 10 per hour. What is the

Firdavs [7]

Answer:are they any answer choice

Step-by-step explanation:

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3 years ago

Identify the transformation that carries the figure onto itself.

kirill [66]

Answer: a

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3 years ago

Read 2 more answers

Suppose that the lifetime of tv tubes are normally distributed with a standard deviation of 1.1 years. Suppose that exactly 20%

IrinaK [193]

Answer:

4.924 years

Step-by-step explanation:

Lets denote X the lifetime of a tv tube (In years). X has distribution $N(\mu, 1.1)$ , with $\mu$ unknown.

We know that P(X < 4) = 0.2. Using this data, we can find the value of $\mu$ throught standarization.

Lets call $Z = \frac{X - \mu}{1.1}$ the standarization of X. Z has distribution N(0,1), and its cummulative function, $\phi$ is tabulated. The values of $\phi$ can be found in the attached file.

$P(X < 4) = P(\frac{X - \mu}{1.1} < \frac{4 - \mu}{1.1}) = P(Z < \frac{4 - \mu}{1.1}) = \phi(\frac{4 - \mu}{1.1}) = 0.2$

The value q such that $\phi (q) = 0.2$ doesnt appear on the table. We can find it by using the symmetry of the normal density function. The opposite of q, -q must verify that $\phi(-q) = 0.8$ , hence -q must be equal to 0.84. Thus, q = -0.84