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4 years ago

6

Going through York Canal, he drives 5 miles in 15 minutes. Later on as he crosses Stover Lake, he drives 30 minutes at the same

average speed. Which statement about the distances is true?

1 answer:

SVETLANKA909090 [29]4 years ago

5 0

When going through York Canal, his speed is 1 mile per 3 minutes.

If going by this speed while crossing Stover Lake, he would drive 10 miles in 30 minutes.

Therefore, he drove 10 miles.

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How do i Evaluate this expression?

timofeeve [1]

$\text{Hey there!}$

$\bf{9.83+8.2(7.01)}$

$\text{You could solved this equation by doing PEMDAS}$

$\text{PEMDAS simply means}\downarrow$

$\text{Parentheses }$

$\text{Exponents}$

$\text{Multiplication}$

$\text{Division}$

$\text{Addition}$

$\text{Subtraction}$

$\text{In your equation you have \bf{MULTIPLICATION \& ADDITION}}$

$\text{Now, that we have every information we need.. lets solve the problem}$

$\text{We have to do \underline{MULITPLICATION} first then \underline{ADDITION} last}$

$\text{9.83 + 8.2(7.01)}$

$\text{8.2(7.01)=57.482}$

$\text{9.83 + 57.482 = 67.312}$

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Sergeu [11.5K]

Answer:

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Step-by-step explanation:

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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

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

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$P(B' \cup A')$

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

So then we can write the probability like this:

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

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

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