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3 years ago

What is a less than -3?

Mathematics

1 answer:

docker41 [41]3 years ago

4 0

-4
-5
-6
-7
-8
-9
-10
-11
etcetera

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There are 6 flavors of candy in a bag. The probability of randomly selecting strawberry is 1/4, lime is 3/16, cherry is 1/8, and

Iteru [2.4K]

The answer is 3/8. You can find a common denominator of 64 among all of the fractions. From this, you can find how many orange candies there are. There is 64 candies, and 40 of them are other flavors, so there's 24 orange candies. You can create the fraction 24/64 and simplify it down to 3/8.

3 0

3 years ago

Read 2 more answers

A.<br> B.<br> C.<br> D.<br> ????????????

qaws [65]

$n^{16}$ and 9 are the squares of, respectively, $n^8$ and 3.

So, using

$a^2-b^2=(a+b)(a-b)$

we have

$n^{16}-9=(n^8+3)(n^8-3)$

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3 years ago

Which is part of the process in the formation of hail pellets? They are tossed up and down in clouds. They lose layers of ice an

kondaur [170]

Answer:

they are tossed up and down in clouds

3 0

3 years ago

2. Sketch each of the following graphs of quadratic functions.

Ira Lisetskai [31]

Hope your able to see it
Have a nice day :)

5 0

3 years ago

Exercise 2.64. On a test there are 20 true-or-false questions. For each problem the student (a) knows the answer with probabilit

ale4655 [162]

Answer:

$\left(p+\dfrac{p}{2}\right)^{19}\left(20-19p-\dfrac{19r}{2}\right)$

Step-by-step explanation:

Let the probability that he gets the correct answer be m.

If he gets the answer, then either he knows it (a probability of p) or he does not know it but guesses correctly (r × 1/2). Therefore,

$m = p + \dfrac{r}{2}$

Let the probability that he gets the incorrect answer be n. Then, by being complementary events

$n=1-m =1 - p - \dfrac{r}{2}$

This is a binomial or Bernoulli distribution since he gets the answer either correctly or incorrectly. The probability of at least 19 correct answers out of 20 is the sum of the probabilities of 19 or 20 correct answers.

$P(\gt19) = P(19) + P(20)$

$P(\gt19) = \binom{20}{19}m^{19}n^1 + \binom{20}{20}m^{20}$

$P(\gt19) = 20\left(p + \dfrac{r}{2}\right)^{19}\left(1 - p - \dfrac{r}{2}\right) + \left(p + \dfrac{r}{2}\right)^{20}$

$= \left( p + \dfrac{r}{2}\right)^{19}\left(20\left(1 - p - \dfrac{r}{2}\right)+p + \dfrac{r}{2}\right)$

$P(\gt19) = \left(p+\dfrac{p}{2}\right)^{19}\left(20-19p-\dfrac{19r}{2}\right)$

3 0

4 years ago