Find the value of the given variable (-4, r) (-8, 3) m=5

How is the way the decimals point moves when you divide a decimal number by a power of ten the same as when you multiply? how is

Svetach [21]

Multiplying decimals is the same as multiplying whole numbers except for the placement of the decimal point in the answer. When you multiply decimals, the decimal point is placed in the product so that the number of decimal places in the product is the sum of the decimal places in the factors.

5 0

3 years ago

Use a table of function values to approximate an x-value in which the exponential function exceeds the polynomial function.f(x)

ycow [4]

X= 2 I believe my friend

5 0

4 years ago

Read 2 more answers

How would I solve I need to know Asap and what is the answer

stealth61 [152]

Answer:

<h2>Here is the correct answer </h2>

(A. 46)

Step-by-step explanation:

STUDY CORRECTION.

3 0

2 years ago

Add parentheses to make true<br> 6 + 4 x 6 + 2 :- 8

d1i1m1o1n [39]

Answer:

62

Step-by-step explanation:

4 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