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2 years ago

Anyone wanna 1v1 on RL?

Mathematics

1 answer:

german2 years ago

7 0

Answer:

in what I play on Xbox, what you tryna play i got 2k21, cod and other games but that the only games I play but I got more if u play Xbox name Tripple k3

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Anthony is an executive assistant. He makes $15.85 per hour. If he worked 47.50 hours this week, what is his (a) OT pay? What ab

alexdok [17]

Anthony's OT pay will be $118.875 and his overall pay will be $752.875.

Number of hours Anthony has to work in a week = 5 × 8 = 40 hours.

He worked for = 47.50 hours.

Overtime = 47.50 hours - 40 hours = 7.50 hours

Pay per hour = $15.85 / hour

Overtime pay (OT pay) = $15.85 × 7.50 = $118.875

Overall pay = $15.85 × 47.50 = $752.875

Therefore, Anthony's OT pay is $118.875 and his overall pay is $752.875.

Learn more about overtime pay here -

brainly.com/question/19022439

#SPJ9

5 0

1 year ago

Given bc=4 cd=11 and bd=10 list angles of triangle bcd least to greatest

Nataly_w [17]

$\bar{B}\bar{C}, \bar{B}\bar{D}, \bar{C}\bar{D}$
$4, 10, 11$

4 0

3 years ago

Please help me somebody! <3

Misha Larkins [42]

I believe it would be -17m-4/8

8 0

2 years ago

Read 2 more answers

Solve the following:<br><br> 1 - .008 =

kykrilka [37]

Answer:

0.992

Step-by-step explanation:

do you not have a calculator

3 0

2 years ago

Read 2 more answers

adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the

Zigmanuir [339]

Answer:

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$ and $\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

Step-by-step explanation:

The equation of the isotope decay is:

$\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }$

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

$\tau = \frac{5568\,years}{\ln 2}$

$\tau \approx 8032.926\,years$

The decay time is:

$t = 1315\,years + 2007\,years \pm 13\,years$ (There is no a year 0 in chronology).

$t = 3335 \pm 13\,years$

Lastly, the relative amount is estimated by direct substitution:

$\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$

$\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

4 0

2 years ago