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eduard
3 years ago
12

Helppp!!!! please!!!

Mathematics
1 answer:
marin [14]3 years ago
8 0

Answer:

b 5.4 square cm

Step-by-step explanation:

Area of figure

=  \frac{1}{2}  \times base \times height \\  \\  =  \frac{1}{2} \times 5.4 \times 2 \\  \\  = 5.4 \:  {cm}^{2}

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How many centimeters are in 35 inches? (1 in. = 2.54 cm)
Colt1911 [192]

Answer:

88.90 cm

Step-by-step explanation:

8 0
3 years ago
Celik, S. (2011). A media comparison study on first aid instruction. Health Education Journal. 72(1), 95-101
aniked [119]

Answer:

The answer is "Option a".

Step-by-step explanation:

In the given scenario, The appropriate test for the teaching profession's accomplishment pre and post-class would be the two-sample t-test with dependent samples, that's why choice "a" is correct.

7 0
3 years ago
1/6 + 3/8 <br> A. 4/6<br> B. 4/8 <br> C. 4/14<br> D. 13/24
Naddika [18.5K]

Here is you're answer:

In order to get you're answer you need to find the common denominator then add.

  • \frac{1}{6} + \frac{3}{8}
  • Find the common denominator:
  • CD = 48
  • = \frac{26}{48}
  • Simplify:
  • \frac{26}{48} \div 2 = \frac{13}{24}
  • = \frac{13}{24}

Therefore you're answer is option D "13/24."

Hope this helps!

8 0
3 years ago
Suppose that 1 kg of Sodium-24 is reduced to 0.0156 kg in 90 hours. What is the half-life, in hours, of Sodium-24?​
N76 [4]

Answer:

14.99 hours

Step-by-step explanation:

The formula for half-life is given as,

R/R' = 2ᵃ/ᵇ...................... Equation 1

Where R= Original mass of sodium-24, R' = mass of Sodium-24 after disintegration, a = Total time taken for sodium-24 to disintegrate, b = half-life of sodiun-24.

Given: R = 1 kg, R' = 0.0156 kg, a = 90 hours

Substitute into equation 1 and solve for b

1/(0.0156) = 2⁹⁰/ᵇ

Taking the log of both side

log[1/(0.0156)] = log(2⁹⁰/ᵇ)

log[1/(0.0156)] = 90/b(log2)

90/b = log[1/(0.0156)]/log2

90/b = 6.0023

b = 90/6.0023

b =  14.99 hours

b = 14.99 hours.

Hence the half-life of Sodium-24 is 14.99 hours

7 0
3 years ago
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
3 years ago
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