What is the justification for 10=15x

hey guys I really need this right now if my grade average is a 75 and I make a 36 on my 9 weeks test what my average be?

nlexa [21]

Answer:

55.5 i think

Step-by-step explanation:

75 + 36 = 111

111/2 = 55.5

7 0

3 years ago

F(x,y)=4x^2y^3-square root 2x+5y in the direction of =-1,4

dedylja [7]

Answer:

frygvhbe 0qcin3ywhjwjhw

Step-by-step explanation:

wbhbjhxw

wjskjwn

djwnkoqn[

W

d

xd

q4444444444

8 0

3 years ago

I need the answer to this problem

Molodets [167]

Answer:

See below

Step-by-step explanation:

When multiplying powers, add them

^3 x^4 = ^7

^2 x ^ 4 =^6

Sorry, putting multiple ones, hard to see in pic

Multiply coefficients

3x3 = 9

Soo . .

9^7 or 9^ 6 depending on the #'s(powers) in the pictures

8 0

3 years ago

The library had 124,000 books find how many books are classified as children fiction l

Mrrafil [7]

Take 124,000 books and multiply it by the percentage of children fiction which is 23%

124,000 x 0.23 = 28,520 children fiction books

6 0

4 years ago

A manufacturing company uses an acceptance scheme on items from a production line before they are shipped. The plan is a two-st

pishuonlain [190]

Answer:

a) The probability that a box containing 3 defectives will be shipped is $51.74\%$

b) The probability that a box containing only 1 defective will be sent back for screening is $17.39\%$

Step-by-step explanation:

Hi

a) The first step is to count the number of total possible random sets of taking a sample size of 4 items over 23 items of the box, so $\left[\begin{array}{ccc}23\\4\end{array}\right] =23C4=8855$

The second step is to count the number of total possible random sets of taking a sample size of 4 items over 20 items of the box (discounting the 3 defectives) as the possible ways to succeed, so $\left[\begin{array}{ccc}20\\4\end{array}\right] =20C4=4845$

Finally we need to compute $\frac{\# ways\ to\ succeed}{\# random\ sets\ of \ 4} =\frac{4845}{8855}=0.5471=P(S)$ , therefore the probability that a box containing 3 defectives will be shipped is $P(S)=54.71\%$

a) The first step is to count the number of total possible random sets of taking a sample size of 4 items over 23 items of the box, so $\left[\begin{array}{ccc}23\\4\end{array}\right] =23C4=8855$

The second step is to count the number of total possible random sets of taking a sample size of 4 items over 22 items of the box (discounting the defective 1) as the possible ways to succeed, so $\left[\begin{array}{ccc}22\\4\end{array}\right] =22C4=7315$

Then we need to compute $\frac{\# ways\ to\ succeed}{\# random\ sets\ of \ 4} =\frac{7315}{8855}=0.8260=P(S)$ , therefore the probability that a box containing 1 defective will be shipped is $P(S)=82.60\%$

Finally the probability that a box containing only 1 defective will be sent back for screening will be $P(BS)=1-P(S)=1-0.8260=0.1739=17.39\%$

4 0

3 years ago