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Jobisdone [24]
3 years ago
14

Find a real number between 7 3/5 and 7 4/5

Mathematics
1 answer:
Sladkaya [172]3 years ago
4 0
7 3/5 = 38/5 = 380/10;

7 4/5 = 39/5 = 390/10;

A real number between 380/10 and 390/10 is 385/10 = 77/2;
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PLEASE HELP ME!!PLEASE HELP ME!!!!!
muminat

Answer:

The actual dimensions would be 3.375 feet by 3.5 feet. The area of the room would be 11.8 square feet.

Step-by-step explanation:

5 0
3 years ago
Find the center and radius of the circle (x+3)^2+(y-1)^2=81
sasho [114]

The equation of a circle is written as ( x-h)^2 + (y-k)^2 = r^2

h and k is the center point of the circle and r is the radius.

In the given equation (x+3)^2 + (y-1)^2 = 81

h = -3

k = 1

r^2 = 81

Take the square root of both sides:

r = 9

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5 0
3 years ago
the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?​
madam [21]

Answer:

About 0.6548 grams will be remaining.  

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

f(t)=a(r)^t

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}

One year has 365 days.

Therefore, the amount remaining after one year will be:

\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548

About 0.6548 grams will be remaining.  

Alternatively, we can use the standard exponential growth/decay function modeled by:

f(t)=Ce^{kt}

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

255=510e^{38k}

Solving for k:

\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)

Thus, our function is:

f(t)=510e^{t\ln(.5)/38}

Then after one year or 365 days, the amount remaining will be about:

f(365)=510e^{365\ln(.5)/38}\approx 0.6548

5 0
3 years ago
For a process, the average range for all samples was 5 and the process average was 25. If the sample size was 10, calculate UCL
alexira [117]

Answer:

8.885

Step-by-step explanation:

Given that :

Sample size, n = 10

The average range, Rbar for all samples = 5

The upper control limit, UCL for the R-chart is :

UCL L= D4Rbar

From the control chart constant table, D4 = 1.777

Hence,

UCL = 1.777 * 5

UCL = 8.885

The UCL for the R-chart is 8.885

7 0
3 years ago
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erastova [34]

Answer:

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Step-by-step explanation:

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