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3 years ago

The rule I(p) = 400p - 20p^2 can be used to estimate the income (I) from ticket sales on based on the ticket price (p)

Mathematics

1 answer:

Lostsunrise [7]3 years ago

3 0

‘Shh d hbd hubs Dugan. Sub sub e. Skins suns

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There are three cardinals, five sparrows and four finches near a bird feeder. If a sharp shunned hawk randomly selects one song

Hoochie [10]

Answer:

1/4

Step-by-step explanation:

The probability of it being a cardinal is the number of cardinals / the total number of possible birds the hawk could eat.

There are 3 + 5 + 4 = 12 total birds near the bird feeder.

There are 3 cardinals.

So the probability is 3/12 = 1/4 that the cardinal is eaten!

8 0

3 years ago

How to multiply a whole number by a fraction

AURORKA [14]

Answer:

There are 3 simple steps to multiply fractions

1. Multiply the top numbers (the numerators).

2. Multiply the bottom numbers (the denominators).

3. Simplify the fraction if needed.

Step-by-step explanation:

can i get brainliest pls?

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3 years ago

A company that produces fine crystal knows from experience that 17% of its goblets have cosmetic flaws and must be classified as

Alex73 [517]

Answer:

0.3891 = 38.91% probability that only one is a second

Step-by-step explanation:

For each globet, there are only two possible outcoes. Either they have cosmetic flaws, or they do not. The probability of a goblet having a cosmetic flaw is independent of other globets. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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

And p is the probability of X happening.

17% of its goblets have cosmetic flaws and must be classified as "seconds."

This means that $p = 0.17$

Among seven randomly selected goblets, how likely is it that only one is a second

This is P(X = 1) when n = 7. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{7,1}.(0.17)^{1}.(0.83)^{6} = 0.3891$

0.3891 = 38.91% probability that only one is a second

7 0

3 years ago

Please help the question is in the picture

Oduvanchick [21]

Answer: C

Step-by-step explanation:

You need to substitute x for 2.5. After that, multiply 101 and 2.5 to get 252.5. Then, add 252.5 and 207.85 to get 460.35.

7 0

3 years ago

Find an angle between 0 and 360 that is coterminal with -60.

Svet_ta [14]

The angle which is coterminal with -60 between 0 and 360 is; 300°.

<h3>Which angle is coterminal with -60?</h3>

Since, it follows that the representation of the angle -60 corresponds to 60° span in the clockwise direction.

Therefore, by going the conventional anticlockwise direction, the angle which is coterminal with -60 is; 300°.