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NARA [144]
2 years ago
9

Will give brainliest if correct

Mathematics
1 answer:
zaharov [31]2 years ago
6 0

Answer:

8x+140=180

8x=40

x=5°

Step-by-step explanation:

Sum of angles in a straight line=180°

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Translate into an inequality.
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5x-1 is greater than or equal to -11
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A test consists of 10 true or false questions. To pass the test a student must answer at least eight questions correctly. If the
forsale [732]
<h3>The probability of  student passing the quiz with  at least 50% of the questions correct is 0.05457.</h3>

Step-by-step explanation:

Here, the total number of T/F question = 10

The minimum answers needed correctly answered = 8

So, student needs to answer at least 8 questions correctly.

Here, the possibility of answering a question correctly  = (\frac{1}{2})   = p = 0.5

Also, the possibility of answering a question wrong  = (\frac{1}{2})  = q = 0.5

Now, to pass he needs to answer 8 or more (  8 , 9 or 10) answers correctly.

P(answering 8 correct answer)  = ^{10}C_8(p)^8(q)^2 = ^{10}C_8(0.5)^8(0.5)^2  = 0.0439

P(answering 9 correct answer)  =  ^{10}C_9(p)^9(q)^1 = ^{10}C_9(0.5)^9(0.5)^1  = 0.0097

P(answering 10 correct answer)  = ^{10}C_{10}(p)^{10}(q)^0 = ^{10}C_{10}(0.5)^{10}(0.5)^0  = 0.00097

So, the total Probability   = P(8) + P(9) + P(10)

= (0.0439) + (0.0097) + (0.00097)

= 0.05457

Hence, the probability that the student passes the quiz  with  at least 8 of the questions correct is 0.05457.

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3 years ago
Find the GCF of the monomials: 72x^3 y^2 and 210x^2y^5
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Answer: The GCF of the monomials is 1.

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The SAT scores have an average of 1200 with a standard deviation of 60. A sample of 36 scores is selected. a) What is the probab
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Answer:

a) 0.0082

b) 0.9987

c) 0.9192

d) 0.5000

e) 1

Step-by-step explanation:

The question is concerned with the mean of a sample.  

From the central limit theorem we have the formula:

z=\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }

a) z=\frac{1224-1200}{\frac{60}{\sqrt{36} } }=2.40

The area to the left of z=2.40 is 0.9918

The area to the right of z=2.40 is 1-0.9918=0.0082

\therefore P(\bar X\:>\:1224)=0.0082

b) z=\frac{1230-1200}{\frac{60}{\sqrt{36} } }=3.00

The area to the left of z=3.00 is 0.9987

\therefore P(\bar X\:

c) The z-value of 1200 is 0

The area to the left of 0 is 0.5

z=\frac{1214-1200}{\frac{60}{\sqrt{36} } }=1.40

The area to the left of z=1.40 is 0.9192

The probability that the sample mean is between 1200 and 1214 is

P(1200\:

d) From c) the probability that the sample mean will be greater than 1200 is 1-0.5000=0.5000

e) z=\frac{73.46-1200}{\frac{60}{\sqrt{36} } }=-112.65

The area to the left of z=-112.65 is 0.

The area to the right of z=-112.65 is 1-0=1

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3 years ago
Type an equation for the following pattern.<br> Need Help!!
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Answer:

all work shown and pictured

4 0
3 years ago
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