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3 years ago

11

Miguel reads about a sea turtle and a manatee. The sea turtle is 3 feet long. The manatee is 4feet times as long as the sea turt

le. How long is the manatee??

1 answer:

xeze [42]3 years ago

3 0

The manatee is 12 ft long

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During a sale at a bookstore, Joseph buys a book at full price. He is given a 50 percent discount on a second book of equal or l

NARA [144]

Total cost of books was 15 + 10 or 25. With the 50 % discount on the $10 book, the total is now $20. The total cost was reduced by (B) 20%.
$5 off the $25 price is a 20% discount.

5 0

3 years ago

what is an equivalent fraction to 3/8 ? Find a common denominator for the pair of fractions. Then, write equivalent fractions

Ann [662]

Answer:

6 1/6

Step-by-step explanation:

A fraction that is equivalent to 3/8 is 6 1/6

5 0

3 years ago

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

Elenna [48]

Answer:

Part a: <em>The probability of no arrivals in a one-minute period is 0.000045.</em>

Part b: <em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>

Part c: <em>The probability of no arrivals in a 15-second is 0.0821.</em>

Part d: <em>The probability of at least one arrival in a 15-second period is 0.9179.</em>

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

$X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)$

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

Substitute the value of λ=10 in the formula as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}$

Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}$

<em>The probability of no arrivals in a one-minute period is 0.000045.</em>

Part b:

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

$\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}$

<em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as $\frac{X}{4}$

$\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}$

That is,

$Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)$

So, the probability mass function of Y is,

$P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots$

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}$

<em>The probability of no arrivals in a 15-second is 0.0821.</em>

Part d:

The probability that there is at least one arrival in a 15-second period is calculated as,

$\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}$

<em>The probability of at least one arrival in a 15-second period is 0.9179.</em>

7 0

3 years ago

Vera ordered a soup, salad, and sandwich to share with her friends. The probabilities of each item being ready

Mkey [24]

Answer:

its

Soup ready first

Sandwich ready first

Salad ready first

Step-by-step explanation:

7 0

3 years ago

M,H, and G have written more than a combined total of 22 articles. H has written 1/4 as many as M. G has written 3/2 as M. Write

Natalka [10]

Missing part of the question:

Write an inequality to determine the number of articles, M could have written for the school newspaper.

Answer:

The inequality: $11M > 88$

The solution: $M > 8$

Step-by-step explanation:

Given

From the question, we have the following parameters:

$M + H + G > 22$

$H = \frac{1}{4}M$

$G = \frac{3}{2}M$

Required

Determine the inequality to solve for M

Substitute the values for H and G in the inequality:

$M + H + G > 22$

$M + \frac{1}{4}M + \frac{3}{2}M > 22$

Multiply through by 4

$4(M + \frac{1}{4}M + \frac{3}{2}M) > 22*4$

$4M + M + 6M > 88$

$11M > 88$

Divide both sides by 11

$M > 8$

4 0

3 years ago