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2 years ago

If UW =9x-9, what is UW in units

Mathematics

1 answer:

mash [69]2 years ago

8 0

Answer:

if UW = 9x-9 then UW is 5 in units

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Need help ASAP please and thank you!

erik [133]

Answer:

I believe that the answer is 153 units^2.

9*10=90

And then 9*7=63, since there are two of the same triangle, we don't have to divide by 2.

So, 90+63=153.

I hope this helps!

7 0

3 years ago

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Help me out plz I need an answer

weeeeeb [17]

The correct answer is B
Hope this helped :)

7 0

3 years ago

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Using descriptive statistics, you have found out that it takes an average of 40 minutes to complete a mid-term examination in IS

choli [55]

Answer:

b. exam completion time is negatively skewed

Step-by-step explanation:

A data distribution is said to be negatively skewed when median value of the distribution is higher than the average value of the distribution.

In this example

average mid-term completion time is 40 minutes

median mid-term completion time is 55 minutes.

thus median > mean, so the mid-term completion time is negatively skewed. Negatively skewed distributions are also called left-skewed.

6 0

3 years ago

What is the degree of the polynomial F(x)=2x^3-x^2+5x-3

Temka [501]

the degree is the largest exponent on the variable, so it would be 3-

3 0

3 years ago

A commuter crosses one of three bridges, A, B, or C, to go home from work. The commuter crosses bridge A with probability 1/3, c

mihalych1998 [28]

Answer:

0.1333 = 13.33% probability that bridge B was used.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Arrives home by 6 pm

Event B: Bridge B used.

Probability of arriving home by 6 pm:

75% of 1/3(Bridge A)

60% of 1/6(Bridge B)

80% of 1/2(Bridge C)

$P(A) = 0.75*\frac{1}{3} + 0.6*\frac{1}{6} + 0.8*\frac{1}{2} = 0.75$

Probability of arriving home by 6 pm using Bridge B:

60% of 1/6. So

$P(A \cap B) = 0.6*\frac{1}{6} = 0.1$

Find the probability that bridge B was used.

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1}{0.75} = 0.1333$

0.1333 = 13.33% probability that bridge B was used.

8 0

3 years ago