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levacccp [35]
3 years ago
11

The average of 3,5,10 and X is 6, what is the value of X?

Mathematics
1 answer:
Rashid [163]3 years ago
6 0
x = 6
The average of 4 numbers is 6, which means that (3 + 5 + 10 + x) all divided by 4 is 6. That means that 3 + 5 + 10 + x = 24, because 24/4 is 6.

18 + x = 24, subtract 18 from 24 to get x = 6.
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Find the mode of the data.
azamat

Answer:

1

Step-by-step explanation:

The mode is the number that appears most often

1 appears twice in the data list and that is the mode

7 0
3 years ago
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Find the simple interest on a $340 loan at a 15% annual interest rate for 1<br> year
Crank

Answer:

$51.

Step-by-step explanation:

Interest I = PRT/100   where P = amount of the loan , R = rate and T = the times in years.

I = 340 * 15 * 1 / 100

= 5100/100

= $51.

4 0
3 years ago
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How do you work this out?
Ira Lisetskai [31]
I'm guessing you find the length of the hypotenuse using the pythagorean theorem. 

a² + b² = c²     a and b are the legs and c is the hypotenuse

15² + 8² = c²

225 + 64 = c²

289 = c²

√289 = √c²

17 = c

hope that helps, God bless!
3 0
4 years ago
Read 2 more answers
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
BRAINLIEST ASAPPPPPPPPP
Natali [406]
The answer is C. The IQR is the best measurement of spread for games and movies
hoped this helped
5 0
4 years ago
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