What's The Domain And Range

Mathematics

1 answer:

Ostrovityanka [42]4 years ago

6 0

The domains are the first ones like 5,2,-3,-1, 7 and range are the second ones

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An eight-sided die, which may or may not be a fair die, has four colors on it; you have been tossing the die for an hour and hav

Ierofanga [76]

Answer:

$P(Yellow) = \frac{29}{147}$

Step-by-step explanation:

Given

$Brown = 27\\Purple = 47\\Yellow = 29\\Green = 44$

Required

$P(Yellow)$ --- next roll to be yellow

This is calculated as:

$P(Yellow) = \frac{Yellow}{Total}$

So, we have:

$P(Yellow) = \frac{29}{27 + 47 + 29 + 44}$

$P(Yellow) = \frac{29}{147}$

7 0

3 years ago

Management selection. A corporation plans to fill 2 different positions forâ€‹ vice-president, Upper V 1 and Upper V 2â€‹, from

Mandarinka [93]

Answer:

45 different ways to fill the positions.

Step-by-step explanation:

Consider the provided information.

Management selection. A corporation plans to fill 2 different positions for vice-president, V₁ and Upper V₂, from administrative officers in 2 of its manufacturing plants. Plant A has 9 officers and plant B has 5.

We need to find the probability that 2 positions be filled if the Upper V₁ position is to be filled from plant A and the Upper V₂, position from plant B?

To fill the post of V₁ from plant A, which has 9 officers, we have 9 ways.

To fill the post of V₂ from plant B, which has 5 officers, we have 5 ways.

Therefore, the total number of ways are: $9\times 5=45$

Hence, 45 different ways to fill the positions.

8 0

4 years ago

Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with μ=10

Tju [1.3M]

Answer: 0.8238

Step-by-step explanation:

Given : Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with $\mu=106$ and $\sigma=15$ .

Let x denotes the scores on a certain intelligence test for children between ages 13 and 15 years.

Then, the proportion of children aged 13 to 15 years old have scores on this test above 92 will be :-

$P(x>92)=1-P(x\leq92)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{92-106}{15})\\\\=1-P(z\leq })\\\\=1-P(z\leq-0.93)=1-(1-P(z\leq0.93))\ \ [\because\ P(Z\leq -z)=1-P(Z\leq z)]\\\\=P(z\leq0.93)=0.8238\ \ [\text{By using z-value table.}]$