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3 years ago

Can someone tell me if this is right thanks

Mathematics

2 answers:

andreyandreev [35.5K]3 years ago

7 0

bearing in mind that an absolute value expression is in effect a piece-wise expression, because it has a ± version.

$\bf 3|x|+7=28\implies 3|x|=21\implies |x|=\cfrac{21}{3}\implies |x|=7\implies \begin{cases} +(x)=7\\ -(x)=7 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ +(x)=7\implies \boxed{x=7}~\hfill -x=7\implies \boxed{x=-7}$

saveliy_v [14]3 years ago

3 0

Yeah because 31x1+7-7=28-7=21

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irinina [24]

Answer:
JKLM is a parallelogram
Explanation:
The slope m of a line through two points (x1,y1) and (x2,y2)is given by the formula:
m=Δy/Δx=y2-y1/x2-x1
So the slopes of the sides of our quadrilateral are:
mJK=(−1)−24−(−3)=−37
mKL=(−5)−(−1)2−4=2
mLM=(−2)−(−5)−5−2=−37
mMJ=2−(−2)(−3)−(−5)=2
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3 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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