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3 years ago

I don’t understand number 4

Mathematics

2 answers:

alexgriva [62]3 years ago

5 0

Angle 2 would also be 75 degrees.
75+75 = 150
150 + 90 = 240
240 + 75 = 316
360 - 316 = 44

Angle 1 = 75 degrees.
Angle 2 = 75 degrees Angle 3 = 150 Angle 4 = 90

kobusy [5.1K]3 years ago

3 0

1.75
2.15
3.105
4.90
these are the answers

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Log2(y)=3log3(u) log5(y)=2log(u) + 3log5(v) Write y in terms of u and v

Gelneren [198K]

It’s attached there

Here

7 0

3 years ago

A student takes an exam containing 1414 multiple choice questions. The probability of choosing a correct answer by knowledgeable

Readme [11.4K]

Answer:

0.0082 = 0.82% probability that he will pass

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the students guesses the correct answer, or he guesses the wrong answer. The probability of guessing the correct answer for a question is independent of other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

In this problem we have that:

$n = 14, p = 0.3$ .

If the student makes knowledgeable guesses, what is the probability that he will pass?

He needs to guess at least 9 answers correctly. So

$P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14)$

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

$P(X = 9) = C_{14,9}.(0.3)^{9}.(0.7)^{5} = 0.0066$

$P(X = 10) = C_{14,10}.(0.3)^{10}.(0.7)^{4} = 0.0014$

$P(X = 11) = C_{14,11}.(0.3)^{11}.(0.7)^{3} = 0.0002$

$P(X = 12) = C_{14,12}.(0.3)^{12}.(0.7)^{2} = 0.000024$

$P(X = 13) = C_{14,13}.(0.3)^{13}.(0.7)^{1} = 0.000002$

$P(X = 14) = C_{14,14}.(0.3)^{14}.(0.7)^{0} \cong 0$

$P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) = 0.0066 + 0.0014 + 0.0002 + 0.000024 + 0.000002 = 0.0082$

0.0082 = 0.82% probability that he will pass

6 0

3 years ago

1/7 ÷ 2 =

e-lub [12.9K]

Hey there!

1/7 ÷ 2

= 1/7 ÷ 2/1

1/7 ÷ 2/1

Revert 2/1 to 1/2

1 × 1 / 7 × 2

1 × 1 = 1 ⬅ your NUMERATOR (top number)

7 × 2 = 14 ⬅ your DENOMINATOR (bottom number)

1(1) / 7(2) = 1/14

Answer: 1/14 ✅

Good luck on your assignment and enjoy your day!

~LoveYourselfFirst:)

5 0

3 years ago

I need help pleaseeeeeeeeeeeeeeeee!!!!!!!!!!!!!!!!!!!!!!!!

liq [111]

Answer:

2/12

Step-by-step explanation:

You have to divide 2/3 by 4 which is the same as multiply 2/3 with 1/4.

This will give you 2/12

3 0

2 years ago

Read 2 more answers

HELP PLEASE MATH!! A company is testing tires for wear and tear. A given tire is said to either pass the test (P) or fail the te

Xelga [282]

Answer:

PPF, PFF

Step-by-step explanation:

There are several ways you can list all the possible combinations. A couple of my favorite are a) use a binary counting sequence; b) use a gray code counting sequence.

Using the first method, the binary numbers 000 to 111 can be listed in numerical order as 000, 001, 010, 011, 100, 101, 110, 111. Letting 0=P and 1=F, the ones missing from your list are the ones in italics in my list.

Using the second method, we change the right-most character, then the middle one, and finally the left-most character so there is one change at a time: 000, 001, 011, 010, 110, 111, 101, 100.

After you have a list of all possible combinations, it is a simple matter to compare the given list to the list of possibilities to see which are missing.

5 0

3 years ago

Read 2 more answers