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2 years ago

Natasha was thinking of a number.

Mathematics

2 answers:

skelet666 [1.2K]2 years ago

8 0

Answer:

28.1

Step-by-step explanation:

You can answer questions like this by working backwards.

To do this we can divide 78.2 by 2 and that gets us 39.1

Now we have to subtract 11 from it and that gets us our final answer:

28.1

satela [25.4K]2 years ago

3 0

Answer:

She was thinking of the number 28.1

Step-by-step explanation:

Since we do not initially know what number Natasha was thinking of we have to start with 78.2. From here we have to do the reverse operations of what Natasha did.

So we would divide 78.2 by 2 which gives us 39.1

Then we minus 11 from that and we get 28.1

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You have eleven books but only have room for six of them on a shelf. How many possible ways can the books be arranged?

Verizon [17]

Answer:

I would say 720

Step-by-step explanation:

one possibility that comes to my head is stack them on top of each other in many ways

8 0

3 years ago

A hollow cylinder, open at one end, is constructed of thin sheet metal. The total external surface area of the cylinder is 192pi

MrRa [10]

S A = r² π + 2 r π h
192 π = r² π + 2 r π h / : π
192 = r² + 2 r h
2 r h = 192 - r²
h = (192 - r² ) / 2 r
V = r² π h = r² π ( (192 - r² ) / 2 r ) =
= π / 2 ( 192 r - r³ )

7 0

3 years ago

The probability that a randomly selected 3-year-old male chipmunk will live to be 4 years old is 0.96516.

mezya [45]

Using the binomial distribution, it is found that there is a:

a) The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

b) The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

c) The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

-----------

For each chipmunk, there are only two possible outcomes. Either they will live to be 4 years old, or they will not. The probability of a chipmunk living is independent of any other chipmunk, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

0.96516 probability of a chipmunk living through the year, thus $p = 0.96516$

Item a:

Two is P(X = 2) when n = 2, thus:

$P(X = 2) = C_{2,2}(0.96516)^2(1-0.96516)^{0} = 0.9315$

The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

Item b:

Six is P(X = 6) when n = 6, then:

$P(X = 6) = C_{6,6}(0.96516)^6(1-0.96516)^{0} = 0.80834$

The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

Item c:

At least one not living is:

$P(X < 6) = 1 - P(X = 6) = 1 - 0.80834 = 0.19166$

The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

A similar problem is given at brainly.com/question/24756209

6 0

3 years ago

Pls help sixowiskosixo

erma4kov [3.2K]

The given question is a quadratic equation and we can use several methods to get the solutions to this question. The solution to the equation are 3/4 and -5/6 and the greater of the two solutions is 3/4

<h3>Quadratic Equation</h3>

Quadratic equation are polynomials with a second degree as it's highest power.

An example of a quadratic equation is

$y = ax^2 + bx + c$

The given quadratic equation is $24x^2 + 2x = 15$

Let's rearrange the equation

$24x^2 + 2x = 15\\24x^2 + 2x - 15 = 0$

This implies that

a = 24
b = 2
c = -15

The equation or formula of quadratic formula is given as

$y = \frac{-b +- \sqrt{b^2 - 4ac} }{2a}$

We can substitute the values into the equation and solve

$y = \frac{-b +- \sqrt{b^2 - 4ac} }{2a}\\y = \frac{-2 +- \sqrt{2^2 -4 * 24 * (-15)} }{2*24} \\y = \frac{-2+-\sqrt{4+1440} }{48} \\y = \frac{-2+-\sqrt{1444} }{48} \\y = \frac{-2+- 38}{48} \\y = \frac{-2+38}{48} \\y = \frac{3}{4}\\ \\or\\y = \frac{-2-38}{48} \\y = \frac{-40}{48} \\y = -\frac{5}{6}$