Wich numbers are rational (r) or irrational (i)? 68,18/42,70.2329,91.154389,5.

One sample of n = 10 scores has a mean of M = 8, and a second sample of n = 5 scores has a mean of M = 2. If the two samples are

schepotkina [342]

Answer:

The mean for the combined sample = 6.

Step-by-step explanation:

We have been given that one sample of n = 10 scores has a mean of M = 8.

So the sum of 10 scores for 1st sample will be: $10\times 8=80$

We are also told that second sample of n = 5 scores has a mean of M = 2.

So the sum of 5 scores for 2nd sample will be: $5\times 2=10$

When the both samples are combined, so total points will be: $80+10=90$ and total scores will be $10+5=15$ .

$\text {Mean for the combined sample}=\frac{\text{Total points of both data sets}}{\text{Total scores of both data sets}}$

$\text {Mean for the combined sample}=\frac{90}{15}$

$\text {Mean for the combined sample}=6$

Therefore, the mean for the combined sample will be 6.

6 0

3 years ago

Find the missing exponent

sladkih [1.3K]

-7 is the missing exponent

3 0

2 years ago

(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

2 years ago

PLEASE HELP ME!!! The sum of two numbers is 46. When 30 is subtracted from 5 times the smaller the result is 3 times the larger.

Ipatiy [6.2K]

Answer:

m = 21

n= 25

Step-by-step explanation:

the equations are:

n + m= 46

5m-30 = 3n

plz vote brainliest :)

8 0

3 years ago

Need your help asap trying to get 100%

marta [7]

Answer:

if he sells 12 every 8 months, supposing the trend continues, in 34 months he’ll sell...

12(properties) = 8 months

12/8 = 1 month

1 month = 1.5(properties)

2 months = 3 properties

(since 1.5 properties is unrealistic, but it helps in the math, to find for one month)

so for 34 months he’ll sell...

1.5*34 = 51 properties

4 0

3 years ago