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4 years ago

What is 9 5/7 as an improper fraction?

Mathematics

1 answer:

andrew11 [14]4 years ago

5 0

Answer:

68/7

Step-by-step explanation:

9*7+5

63+5

68/7

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Enter the perimeter of a 30°-60°-90° triangle with a hypotenuse of 20. Round your answer to the nearest tenth. You can leave the

motikmotik

Answer:

47.3

Step-by-step explanation:

as shown above ,this question can be solved using trigonometry and hence we get the perimeter to be 47.3

3 0

3 years ago

Please help confused with half life /compound interest problems

jekas [21]

Half life means only half of matter is left, which means half is decayed/radiated.
Formula for half life is
t/2 = [(elapsed time)*(log2)]/[log(beginning amount/ending amount)]
1. t/2 = [(10)*(log2)]/[log(100/76.1)] = 25.378 minutes.
t/2 = [10 * 0.301]/[log 1.314] = [3.01]/[0.1186] = 25.378
2. 53.12 = [(T)*(log2)]/[log(15/4)], on solving we get, T = 101.29 days
53.12 = [T * 0.301]/[log 3.75] = [0.301 * T]/[0.574],
T = [53.12 * 0.574]/[0.301] = [30.49]/[0.301] = 101.29<span />

6 0

3 years ago

One sample has Aa Aa One sample has n 10 scores and a variance of s2 20, and a second sample has n 15 scores and a variance of s

siniylev [52]

Answer:

option (a) It will be closer to 30 than to 20

Step-by-step explanation:

Data provided in the question:

For sample 1:

n₁ = 10

variance, s₁² = 20

For sample 2:

n₂ = 15

variance, s₂² = 30

Now,

The pooled variance is calculated using the formula,

$S^{2}_{p} = \frac{(n_{1}-1)\times s^{2}_{1} +(n_{2}-1)\times s^{2}_{2}}{n_{1}+n_{2}-2}$

on substituting the given respective values, we get

$S^{2}_{p} = \frac{(10-1)\times 20 +(15-1)\times 30}{10+15-2}$

$S^{2}_{p}$ = 26.0869

Hence,

the pooled variance will be closer to 30 than to 20

Therefore,

The correct answer is option (a) It will be closer to 30 than to 20

4 0

3 years ago

Please help.. What is the surface area of the rectangular prism below?

In-s [12.5K]

Answer:

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Find the length of the curve x=e^t e^{-t},\;\;y=5-2t,\;\;0 \le t \le 3.

Oksana_A [137]

$\begin{cases}x(t)=e^t+e^{-t}\\y(t)=5-2t\end{cases}\implies\begin{cases}x'(t)=e^t-e^{-t}\\y'(t)=-2\end{cases}$

The length of the curve is given by the integral

$\displaystyle\int_0^3\sqrt{(e^t-e^{-t})^2+(-2)^2}\,\mathrm dt$

Expand and rewrite the integrand:

$(e^t-e^{-t})^2+(-2)^2=e^{2t}+2+e^{-2t}$
$=e^{-2t}(e^{4t}+2e^{2t}+1)$
$=e^{-2t}(e^{2t}+1)^2$
$\implies\sqrt{(e^t-e^{-t})^2+(-2)^2}=\dfrac{e^{2t}+1}{e^t}$

Now the integral is

$\displaystyle\int_0^3\dfrac{e^{2t}+1}{e^t}\,\mathrm dt=\int_0^3(e^t+e^{-t})\,\mathrm dt$
$=2\displaystyle\int_0^3\cosh t\,\mathrm dt$
$=2\sinh t\bigg|_{t=0}^{t=3}$
$=2\sinh 3$

7 0

3 years ago