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2 years ago

Solve for b. -1lb + 7 = 40

Mathematics

1 answer:

MariettaO [177]2 years ago

7 0

Answer:

b = -33/l

Step-by-step explanation:

-1lb + 7 = 40

-1lb = 33 (subtraction property of equality)

-1lb/l = 33/l (division property of equality)

-1b/-1 = -33/l (division property of equality)

b = -33/l (solved)

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irinina [24]

Answer:

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Step-by-step explanation:

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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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2 years ago

Find the derivative of 3e^y+x=y

Sindrei [870]

You have the following function:

$y=3e^y+x$

Derivate implictly the previous expression, as follow:

$y^{\prime}=3e^yy^{\prime}+1$

Where you have used that:

$(e^y)^{\prime}=e^yy^{\prime}$

Then, the implicit derivative of the given expression is:

$y^{\prime}=3e^yy^{\prime}+1$

Next, solve for y' as follow:

$\begin{gathered} y^{\prime}-3e^yy^{\prime}=1 \\ (1-3e^y)y^{\prime}=1 \\ y^{\prime}=\frac{1}{1-3e^y} \end{gathered}$

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5 months ago

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sp2606 [1]

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2 years ago

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Svet_ta [14]

Answer:

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1 year ago

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