Do these pair of values (x and y) represent two quantities that are proportional

PLEASE HELP ASAP WILLL MARK BRANLIEST TO WHOEVER ANSWERS CORRECTLY FIRST 20 PTS

DedPeter [7]

Answer:

4

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

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:

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

5 0

2 years ago

How to solve and answer using elimination

grin007 [14]

Answer:

the answer is:

x=6

y=7

4 0

3 years ago

The base of a cab in New York is $2.50 as soon as you enter the cab. There is an additional charge of $2.50 per mile traveled. C

Anastaziya [24]

Answer:

7 miles

Step-by-step explanation:

3 0

3 years ago

Scores on a university exam are Normally distributed with a mean of 78 and a standard deviation of 8. The professor teaching the

mafiozo [28]

Answer:

Porcentage of students score below 62 is close to 0,08%

Step-by-step explanation:

The rule

68-95-99.7

establishes:

The intervals:

[ μ₀ - 0,5σ , μ₀ + 0,5σ] contains 68.3 % of all the values of the population

[ μ₀ - σ , μ₀ + σ] contains 95.4 % of all the values of the population

[ μ₀ - 1,5σ , μ₀ + 1,5σ] contains 99.7 % of all the values of the population

In our case such intervals become

[ μ₀ - 0,5σ , μ₀ + 0,5σ] ⇒ [ 78 - (0,5)*8 , 78 + (0,5)*8 ] ⇒[ 74 , 82]

[ μ₀ - σ , μ₀ + σ] ⇒ [ 78 - 8 , 78 +8 ] ⇒ [ 70 , 86 ]

[ μ₀ - 1,5σ , μ₀ + 1,5 σ] ⇒ [ 78 - 12 , 78 + 12 ] ⇒ [ 66 , 90 ]

Therefore the last interval

[ μ₀ - 1,5σ , μ₀ + 1,5 σ] ⇒ [ 66 , 90 ]

has as lower limit 66 and contains 99.7 % of population, according to that the porcentage of students score below 62 is very small, minor than 0,15 %

100 - 99,7 = 0,3 %

Only 0,3 % of population is out of μ₀ ± 1,5 σ, and by symmetry 0,3 /2 = 0,15 % is below the lower limit, 62 is even far from 66 so we can estimate, that the porcentage of students score below 62 is under 0,08 %

6 0

2 years ago