Find the diameter of M. round to nearest hundredth

The second one please help I do not understand and if you can please explain

Masja [62]

When a temperature change causes a substance to turn from liquid to sold, we call the process as FREEZING.

For example, if we take a glass of water and keep it in the freezer, water becomes ice in few minutes.

Since the temperature decreases, water freezes and it becomes ice.

In Ice cream shops, they preserve ice creams in freezer box to prevent them from melting.

6 0

3 years ago

Complete the multiplication: 2EA

Lady bird [3.3K]

In mathematics, a matrix (plural matrices) is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns.

Given that

E = [ 1 2]

A = | 3 0|

| 2 -1 |

2EA = 2 [1 2] | 3 0|

| 2 -1 |

2EA = [ 14 -4]

8 0

4 years ago

Read 2 more answers

there are 45 new houses built. Last month 2/3 of them sold. This month, 1/5 of the remaining houses were sold. How many houses l

Nady [450]

Answer: 9

Step-by-step explanation: 45 times 2/3 is 30. So, 30 houses were sold last month. 30 times 1/5 is 6. So, 6 houses were sold this month. Add them together. 30+6=36. Subtract that from 45 and you will get 9. There are 9 houses left to be sold.

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%.

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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

Identify the slope and y-intercept of the graph that represents each linear situation and interpret what

KIM [24]

Charlidamelio is overrated

4 0

3 years ago