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3 years ago

How to solve for literal equations with multiple variables

1 answer:

6 0

See examples for the method to solving literal equations for a given variable: Solve A = bh for b. Since h is multiplied times b, you must divide both sides by h in order to isolate b. Since (c+d) is divided by 2, you must first multiply both sides of the equation by 2

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The number of accidents that occur at a busy intersection is Poisson distributed with a mean of 4.5 per week. Find the probabili

Mariana [72]

Answer:

a) 0.01111

b) 0.4679

c) 0.33747

Step-by-step explanation:

We are given the following in the question:

The number of accidents per week can be treated as a Poisson distribution.

Mean number of accidents per week = 4.5

$\lambda= 4.5$

Formula:

$P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}$

a) No accidents occur in one week.

$P(x =0)\\\\= \displaystyle\frac{4.5^0 e^{-4.5}}{0!}= 0.01111$

b) 5 or more accidents occur in a week.

$P( x \geq 5) = 1-\displaystyle \sum P(x$

c) One accident occurs today.

The mean number of accidents per day is given by

$\lambda = \dfrac{4.5}{7} = 0.64$

$P(x =1)\\\\= \displaystyle\frac{0.64^1 e^{-0.64}}{1!}= 0.33747$

5 0

3 years ago

5 (RootIndex 3 StartRoot x EndRoot) + 9 (RootIndex 3 StartRoot x EndRoot)

Ratling [72]

Answer:

C. 14 RootIndex 3 StartRoot x EndRoot)

Step-by-step explanation:

To obtain the sum ; the a clear explanation has been attached in the picture below ;

8 0

3 years ago

A coin is tossed and a single 6 sided number cube is rolled. Find the probability of landing on the heads side of the coin and r

Delicious77 [7]

Answer:

The probability of landing on the heads side of the coin and rolling 3 on the number cube is 0.083.

Step-by-step explanation:

Given:

Coin is tossed and a single 6 sided number cube is rolled.

Let the event of tossing head be 'H' and event of rolling a 3 be 'R3'.

Now, we know that:

Probability of an event 'A' = Favorable outcomes of 'A' ÷ Total possible outcomes

Now, for tossing a coin, the total possible outcomes are head and tail. So, there are 2 possible outcomes.

So, probability of event 'H' is given as:

$P(H)=\dfrac{n(H)}{n(S)}\\\\P(H)=\frac{1}{2}=0.5$

Similarly, for rolling a 6 sided cube, the possible outcomes are numbers 1 to 6. So, the number of possible outcomes is, $n(S)=6$

Now, probability of rolling a 3 is given as:

$P(R3)=\frac{n(R3)}{n(S)}\\\\P(R3)=\frac{1}{6}$

Now, both the events 'H' and 'R3' occur together. So, the combined probability is the product of two individual probabilities as they are independent events. So,

$P(H\ and\ R3)=P(H)\times P(R3)\\\\P(H\ and\ R3)=\frac{1}{2}\times \frac{1}{6}\\\\P(H\ and\ R3)=\frac{1}{12}=0.083$

Note: Independent events are those events whose intersection is an empty set of events or the outcome of one event doesn't affect the outcome of the other event.

Therefore, the probability of landing on the heads side of the coin and rolling 3 on the number cube is 0.083.

4 0

4 years ago

You cannot tessellate six-sided regular polygons by themselves. A. True B. False

Nataly_w [17]

Answer:

true

Step-by-step explanation:

baaaaam

5 0

4 years ago

Read 2 more answers