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4 years ago

6 x (8 - 3) = (? x 8) - (? x 3) what are the answers for the question marks?

Mathematics

2 answers:

soldier1979 [14.2K]4 years ago

7 0

The answer is 6.

THey are using distributive property.

6 x (8-3) = 30. 6x5=30.

6 x 8 equals 48. 6 x 3 equals 18. 48-18 equals 30.

Please tell me if I am wrong, and hope it helps!

Elina [12.6K]4 years ago

3 0

6 x (8 - 3) =

(6 x 8) - (6 x 3)

The answer is 6

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Nostrana [21]

Answer:

Considering is a compunded increase rate of the yearly car insurance fee and x is the initial value of the insurance fee then

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4 years ago

Given the data set (18,14, 12, 14, 11, 11, 19, 20, 16, 11), which statements are correct?

Anuta_ua [19.1K]

The correct statement is that the <u>range of the data is 9</u>

<h3>Range of a data</h3>

The formula for calculating the range of data is expressed as:

Range = Highest value - Lowest value

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Learn more on range here: brainly.com/question/2264373

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2 years ago

A bank in the Bay area is considering a training program for its staff. The probability that a new training program will increas

WITCHER [35]

Answer:

$P(B' \cup A') = P((A \cap B)') = 1-P(A \cap B)= 1-0.32=0.68$

See explanation below.

Step-by-step explanation:

For this case we define first some notation:

A= A new training program will increase customer satisfaction ratings

B= The training program can be kept within the original budget allocation

And for these two events we have defined the following probabilities

$P(A) = 0.8, P(B) = 0.2$

We are assuming that the two events are independent so then we have the following propert:

$P(A \cap B ) = P(A) * P(B)$

And we want to find the probability that the cost of the training program is not kept within budget or the training program will not increase the customer ratings so then if we use symbols we want to find:

$P(B' \cup A')$

And using the De Morgan laws we know that:

$(A \cap B)' = A' \cup B'$

So then we can write the probability like this:

$P(B' \cup A') = P((A \cap B)')$

And using the complement rule we can do this:

$P(B' \cup A') = P((A \cap B)')= 1-P(A \cap B)$

Since A and B are independent we have:

$P(A \cap B )=P(A)*P(B) =(0.8*0.4) =0.32$

And then our final answer would be:

$P(B' \cup A') = P((A \cap B)') = 1-P(A \cap B)= 1-0.32=0.68$

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