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3 years ago

HELP Please!!! I need this done quick.

Mathematics

2 answers:

Crazy boy [7]3 years ago

7 0

Hey ok so for A the answer is 1.14 and for b it’s 0.9 hope this helps

matrenka [14]3 years ago

6 0

What form do you need it in?

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neonofarm [45]

Answer:

the answsr is 144 g

Step-by-step explanation:

mass of grapes=total mass-mass of bowl

=265-121

=144 grams

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3 years ago

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zheka24 [161]

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3 years ago

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-12 times 1over3 help.

sergey [27]

Answer:

-4

Step-by-step explanation:

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3 years ago

2) State in your own words, what relationship could be observed between the total

Alex17521 [72]

Answer:

Newton's second law of motion describes the relationship between force and acceleration. They are directly proportional. If you increase the force applied to an object, the acceleration of that object increases by the same factor. In short, force equals mass times acceleration.

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3 years ago

Car colors preference change over the years and according to the particular model that the customer selects. In a recent year, 1

Margarita [4]

Answer:

99.05% probabilitiy that at most six cars are black.

Step-by-step explanation:

For each luxury car, there are only two possible outcomes. Either it is black, or it is not black. The probability of a luxury car being black is independent from other luxury cars. 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.

In this problem we have that:

$p = 0.1, n = 25$

If 25 cars of that year and type are randomly selected, find the probabilitiy that at most six cars are black.

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

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

$P(X = 0) = C_{25,0}.(0.1)^{0}.(0.9)^{25} = 0.0718$

$P(X = 1) = C_{25,1}.(0.1)^{1}.(0.9)^{24} = 0.1994$

$P(X = 2) = C_{25,2}.(0.1)^{2}.(0.9)^{23} = 0.2659$

$P(X = 3) = C_{25,3}.(0.1)^{3}.(0.9)^{22} = 0.2265$

$P(X = 4) = C_{25,4}.(0.1)^{4}.(0.9)^{21} = 0.1384$

$P(X = 5) = C_{25,5}.(0.1)^{5}.(0.9)^{20} = 0.0646$

$P(X = 6) = C_{25,6}.(0.1)^{6}.(0.9)^{19} = 0.0239$

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0718 + 0.1994 + 0.2659 + 0.2265 + 0.1384 + 0.0646 + 0.0239 = 0.9905$

99.05% probabilitiy that at most six cars are black.

7 0

4 years ago