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3 years ago

Dose anyone know how to solve this?

Mathematics

1 answer:

Ghella [55]3 years ago

8 0

Answer:

Domain = [-5, -2) U (1, 3)

Range = (-3, 4]

Step-by-step explanation:

Domain is the x values

Domain = [-5, -2) U (1, 3)

Range is the y values

Range = (-3, 4]

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A tree is currently 8 feet tall and growing at a rate of 3 feet per year. How tall will the tree be in 12 years?

aleksandr82 [10.1K]

Answer:

Step-by-step explanation:

3 x 12 =36

36+8 =44

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2 years ago

Read 2 more answers

Question 6 The mineral content of a particular brand of supplement pills is normally distributed with mean 490 mg and variance o

AysviL [449]

Answer:

0.3085 = 30.85% probability that a randomly selected pill contains at least 500 mg of minerals

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean 490 mg and variance of 400.

This means that $\mu = 490, \sigma = \sqrt{400} = 20$

What is the probability that a randomly selected pill contains at least 500 mg of minerals?

This is 1 subtracted by the p-value of Z when X = 500. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{500 - 490}{20}$

$Z = 0.5$

$Z = 0.5$ has a p-value of 0.6915.

1 - 0.6915 = 0.3085

0.3085 = 30.85% probability that a randomly selected pill contains at least 500 mg of minerals

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3 years ago

The Histogram is a graphic which is: A. Bar Chart useful for showing the distribution of data B. 80-20 principle C. A complex ti

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Step-by-step explanation: Hope this helped :)

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3 years ago

Find the number if 5/3% of it is 4.75

Lostsunrise [7]

If 5/3% of some number is 4.75, then 1/3% of that number would be 4.75/5. Then multiply that answer by 3 to get 3/3=1% of the number, and finally multiply by 100 to get 100%.

7 0

3 years ago

A bridge hand is made up of 13 cards from a deck of 52. find the probability that a hand chosen at random contains at least 3 ki

Vlad [161]

1.
In total there are C(52, 13) ways that we can pick a hand, that is $\frac{52!}{13!39!}$

2.
P(a hand contains at least 3 kings)
=P(a hand contains exactly 3 kings)+P(a hand contains 4 kings)

3.
first let's find P(a hand contains exactly 3 kings):

P(a hand contains exactly 3 kings)
=n(a hand contains exactly 3 kings)/C(52, 13)

n(a hand contains exactly 3 kings)=C(4, 3)*C(48, 10)

where C(4,3) is the total number of ways we can pick 3 out of 4 kings,

C(48, 10) is the number of picking 10 letters to complete a hand, out of the 52-4=48 non-king cards.

so P(a hand contains exactly 3 kings)=[C(4, 3)*C(48, 10)]/C(52, 13)

4. with the same reasoning as in step 3:

P(a hand contains 4 kings)=n(a hand contains 4 kings)/C(52, 13)

= [C(4, 4)*C(48, 9]/C(52, 13)

5.

P(a hand contains at least 3 kings)
=P(a hand contains exactly 3 kings)+P(a hand contains 4 kings)

=[C(4, 3)*C(48, 10)]/C(52, 13)+ [C(4, 4)*C(48, 9)]/C(52, 13)

= $\frac{C(4, 3)*C(48, 10)+C(4, 4)*C(48, 9)}{C(52, 13)}$

= $\frac{4* \frac{48!}{10!38!} + \frac{48!}{9!39!}}{ \frac{52!}{13!39!} }$

simplify by 38! in the denominators and 48! in the numerators :

$\frac{4* \frac{1}{10!} + \frac{1}{9!39}}{ \frac{52*51*50*49}{13!39} } 

$

now simplify by 9! in all denominators:

$\frac{ \frac{4}{10}+ \frac{1}{39} }{ \frac{52*51*50*49}{13*12*11*10*39}}$

$\frac{ 0.426 }{ 9.7} =0.044$

4 0

3 years ago