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Wittaler [7]
3 years ago
9

10yd 4yd what is the area of the shaded portion to the nearest hundredth l

Mathematics
1 answer:
WARRIOR [948]3 years ago
7 0

Answer:

I don't understand the question. Pls is there any picture

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A TV studio has brought in 8 boy kittens and 10 girl kittens for a cat food commercial.
kondor19780726 [428]

Answer:

0.323 = 32.3% probability that the director chooses 3 boy kittens and 5 girl kittens.

Step-by-step explanation:

The kittens are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

A TV studio has brought in 8 boy kittens and 10 girl kittens for a cat food commercial.

This means that N = 8 + 10 = 18

We want 3 boys, so k = 8

The director is going to choose 8 of these kittens at random to be in the commercial.

This means that n = 8

What is the probability that the director chooses 3 boy kittens and 5 girl kittens?

This is P(X = 3).

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 3) = h(3,18,8,8) = \frac{C_{8,3}*C_{10,5}}{C_{18,8}} = 0.323

0.323 = 32.3% probability that the director chooses 3 boy kittens and 5 girl kittens.

3 0
3 years ago
HELP!
dedylja [7]
15+12+8+x=44
35-44=9
x=9
7 0
4 years ago
Read 2 more answers
Solve for f<br> −f+2+4f=8−3f
Amiraneli [1.4K]

Answer:

f=1

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Please solve number 16 for me.
vodka [1.7K]

Answer:

post pictures also for your answer

4 0
3 years ago
A factory produces widgets whose length must be within 1.5 mm of an ideal length of 47 mm. A factory supervisor wants to mark ru
Nat2105 [25]

Using absolute value, it is found that:

  • The situation is described by the following equation: |l - 47| = 1.5
  • The markings should be between 45.5mm and 48.5mm.

---------------------------

The definition of the <em>absolute value function</em> is as follows:

|x| = x, x \geq 0

|x| = -x, x < 0

----------------------------

The length should be set to 47mm with an allowance of 1.5mm, which means that <u>the absolute value of the difference between the length and 47 should be 1.5</u>, thus:

|l - 47| = 1.5

----------------------------

The solutions are:

l - 47 = -1.5 \rightarrow l = 45.5

l - 47 = 1.5 \rightarrow l = 48.5

The markings should be between 45.5mm and 48.5mm.

A similar problem is given at brainly.com/question/24514895

3 0
3 years ago
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