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3 years ago

Please hurry please

Mathematics

2 answers:

Liula [17]3 years ago

8 0

M=102
102/51= 2
2< or equal to 2

Softa [21]3 years ago

6 0

Answer:

102

Step-by-step explanation:

You might be interested in

What is 2 8/10 + (-3.5) in decimal form, Show your work

SSSSS [86.1K]

<h2>QUESTION:</h2>

What is 2 8/10 + (-3.5) in decimal form.

<h2>SOLUTION:</h2>

▶️ 28/10 - 3.5

▶️ 2.8 - 3.5

▶️ - 0.7

<h2>ANSWER:</h2>

- 0.7 ( Minus zero point seven )

<h2 />

7 0

3 years ago

Use the drop-down menus to locate integers on the number line. A B C D 5 -4 3 2 - 1 + 3 4 1 2 5 6 7 B В C < D

vichka [17]

Answer:

can you take a picture of it? the format you put it in is confusing.

5 0

3 years ago

Find the size of each of the angles marked with letter

GuDViN [60]

Answer:

angle l = 115

angle k = 115

angle j = 114

angle i = 66

Step-by-step explanation:

angle l = it is opposite to 115 so its the same

angle k = opposite opposing angles with angle l

angle j = 180-66=114 (between 2 parallel lines the sum of 2 angles is 180)

andle i = 180-114=66 (straight line = 180 degrees so subtract angle j to get i)

4 0

1 year ago

The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

8 0

3 years ago

There are 4 boys and 2 girls from the seventh grade and 3 boys and 5 girls from the eighth grade on the soccer team. Coach Hart

tatuchka [14]

Answer:

The answer is a). 5/24.

Step-by-step explanation:

For the seventh graders there are 4 Boys + 2 Girls which totals 6 students.

The probability that a girl will be picked from the <em>seventh graders is 2/6</em>

For the eighth graders, there are 3 Boys + 5 Girls which totals 8 students

The probability that a girl will be picked from the <em>eighth graders is 5/8</em>

To find the<u> probability that both are girls</u>, we multiply the two fractions together:

2/6 * 5/8 = 5/24

7 0

3 years ago