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3 years ago

List the partial products of 42 x 28

Mathematics

2 answers:

Otrada [13]3 years ago

3 0

The partial products are: 42 * 28 = (40 + 2)*(20 + 8) = 40*20 + 40*8 + 2*20 + 2*8
= 800 + 320 + 40 + 16

MrMuchimi3 years ago

3 0

Answer:

The partial product of $42\times 28$ are 800,320,40,16 and their sum is 1176.

Step-by-step explanation:

The given expression is,

$42\times 28$

In number 42 the place value of 4 is 40 and the place value of 2 is 2. In number 28 the place value of 2 is 20 and the place value of 8 is 8.

To find the partial product, we have multiply the place values of first number with the place values of second number.

Place values of 42 are 40 and 2. Place values of 28 are 20 and 8.

$40\times 20=800$

$40\times 8=320$

$2\times 20=40$

$2\times 8=16$

Therefore the partial product of $42\times 28$ are 800,320,40 and 16.

To find the total product, we have add all partial products.

$T=800+320+40+16=1176$

The value of given expression is 1176.

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Eight percent of all college graduates hired by companies stay with the same company for more than five years. The probability,

Maurinko [17]

Answer:

0.1662 = 16.62% probability that exactly 2 will stay with the same company for more than five years

Step-by-step explanation:

For each graduate, there are only two possible outcomes. Either they stay for the same company for more than five years, or they do not. Graduates are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Eight percent of all college graduates hired by companies stay with the same company for more than five years.

This means that $p = 0.08$

Sample of 11 college graduates:

This means that $n = 11$

The probability, rounded to four decimal places, that in a random sample of 11 such college graduates hired recently by companies, exactly 2 will stay with the same company for more than five years is: the absolute tolerance is +-0.0001

The tolerance means that the answer is rounded to four decimal places.

The probability is P(X = 2). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{11,2}.(0.08)^{2}.(0.92)^{11} = 0.1662$