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inysia [295]
3 years ago
10

Find the area and perimeter for number 19 and 20 with work please

Mathematics
1 answer:
allsm [11]3 years ago
8 0
I promise if you use the internet it will come up
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Fie O centrul unui cerc si doua unghiuri la centru adiacente AOB si BOC.Stiind ca masura unghiului AOB = 120 si masura BOC= 150,
kodGreya [7K]

<span>Let A be the center of a circle and two angles at the adjacent center AOB and BOC. Knowing the measure of the angle AOB = 120 and the measure BOC = 150, find the measures of the angles of the ABC triangle.
</span>solution

Given the above information;
AC=AB, therefore ABC is an isosceles triangle. 
therefore, BAO=ABO=(180-120)/2=30
OAC=OCA=(180-90)/2=45
OBC=BCO=(180-150)/2=15
THUS;
BAC=BAO+OAC=45+30=75
ABC=OBA+CBO=15+30=45
ACB=ACO+BCO=15+45=60


4 0
3 years ago
Please help will mark brainliest
svet-max [94.6K]

Answer:

The answer is 36.9

Step-by-step explanation:

5 0
3 years ago
Click on the measurement that is equal to 20 liters.
Sphinxa [80]

Answer:

A.

Step-by-step explanation:

20000milliters = 20litres

4 0
3 years ago
Read 2 more answers
Jillian’s school is selling tickets for a play. The tickets cost $10.50 for adults and $3.75 for students. The ticket sales for
-Dominant- [34]

Answer:

168

Step-by-step explanation:

The first equation given is 0.50 a +3.75 b= 2071.50

Where a is number of adult tickets and b is number of student tickets

It is also given that 82 students attended, so we can put "82" in place of "b" and then solve the equation for a:

10.50a+3.75b=2071.50\\10.50a+3.75(82)=2071.50\\10.50a+307.5=2071.50\\10.50a=2071.50-307.5\\10.50a=1764\\a=\frac{1764}{10.50}\\a=168

Hence, 168 adults attended

3 0
4 years ago
Read 2 more answers
Consider the probability that at least 88 out of 158 software users will not call technical support. Assume the probability that
Nookie1986 [14]

Answer:

E(X) = 158*0.57 =90.06

And the standard deviation for the random variable is given by:

\sigma= \sqrt{158*0.57*(1-0.57)} = 6.223

And the distribution for the approximation is given by:

X \sim N (\mu = 90.06, \sigma = 6.223)

We can use the z score formula given by:

z = \frac{x -\mu}{\sigma}

And using the complement rule we got:

P(X \geq 88) = 1-P(X

And using the standard normal table we got:

P(X \geq 88) = 1-P(X

Step-by-step explanation:

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Solution to the problem

Let X the random variable of interest, on this case we now that:  

X \sim Bin (n = 158, p =0.57)

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

We want this probability:

P(X\geq 88)

Normal approximation

We need to check if we can use the normal approximation , the conditions are:

np =158*0.57 = 90.06>10

n(1-p) = 158*(1-0.57) = 67.94>10

Since we satisfy both conditions the normal approximation makes sense

The expected value is given by this formula:

E(X) = 158*0.57 =90.06

And the standard deviation for the random variable is given by:

\sigma= \sqrt{158*0.57*(1-0.57)} = 6.223

And the distribution for the approximation is given by:

X \sim N (\mu = 90.06, \sigma = 6.223)

We can use the z score formula given by:

z = \frac{x -\mu}{\sigma}

And using the complement rule we got:

P(X \geq 88) = 1-P(X

And using the standard normal table we got:

P(X \geq 88) = 1-P(X

4 0
3 years ago
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