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2 years ago

PLEASE HELP ME. NO PHONY ANSWERS TY, ANSWER IF YOUR 100% RIGHT!

Mathematics

1 answer:

Ksivusya [100]2 years ago

8 0

The relative frequency of female mathematics majors will be 0.5142.

<h3>How to find the relative frequency?</h3>

The proportion of the examined subgroup's value to the overall account is known as relative frequency.

A sample of 317 students at a university is surveyed.

The students are classified according to gender (“female” or “male”).

The table is given below.

Then the relative frequency of female mathematics majors will be

⇒ 36 / (36 + 34)

⇒ 36 / 70

⇒ 0.5142

Learn more about conditional relative frequency here:

brainly.com/question/8358304

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How did my teacher get 11 for #17

wel

Complementary angles ⇒ the angles add up to 90°

∠A and ∠B are complementary ⇒ ∠A + ∠B = 90°

∠A + ∠B = 90°

Plug in the values for ∠A and ∠B:
(3x - 8) + (5x + 10) = 90

Open the brackets:
3x - 8 + 5x + 10 = 90

Combine like terms:
8x + 2 = 90

Subtract 2 from both sides:
8x = 88

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Answer: x = 11

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3 years ago

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The solution for the following system of linear equation 3m-2n=13 is (2,-1) true or false

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Step-by-step explanation:

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3 years ago

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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.

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3 years ago

jim is planning a party. he spends $150 on a room for the party. he also spends $16.50 per person for food. jim finds the total

meriva

$577.50is the amount he paid.

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3 years ago

Does anybody know how to solve this ?

KATRIN_1 [288]

Answer:

try making a porportion with 57 and 51

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