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SVEN [57.7K]
3 years ago
14

Please help me answer Question 1, and 3! Thanks!!!

Mathematics
1 answer:
Gnom [1K]3 years ago
6 0
I hope this helps you

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Complete the table of values for y=6-2x<br><br> Also please work out 6-2x=3
Whitepunk [10]

Answer:

Step-by-step explanation:

x=0     y=6-2*0=6

x=1      y=6-2*1=4

x=2      y=6-2*2=2

x=3      y=6-2*3=0

6-2x=3

-6       -6

-2x=-3

2x=3

:2    :2

x=3/2

x=1.5

3 0
4 years ago
Solve for m. -5&lt;3+m
Firlakuza [10]

-5 < 3 + m      <em>subtract 3 from both sides</em>

-5 - 3 < 3 - 3 + m

-8 < m → m > -8

8 0
4 years ago
Read 2 more answers
Manuel's dog eats 2/4 bag of dog food in 1 month. His large dog eats 3/4 of dog food in 1 month. How many bags do both dogs eat
Ainat [17]

Answer:

7.5 bags of dog food in 6 months

Step-by-step explanation:

(0.5)(6) = 3 bags for the his smaller dog (0.75)(6) = 4.5 bags for his larger dog... now add 3 + 4.5 and you get 7.5 total bags

8 0
3 years ago
Historically 80% of the monitors made by your company will pass your stringent quality control checks. 40 monitors have just com
madam [21]

Answer:

a) 10.75% probability that exactly 30 of them will pass the quality control checks.

b) Expected value is 32

Variance is 6.4

Step-by-step explanation:

For each monitor, there are only two possible outcomes. Either they will pass the quality checks, or they will not pass these checks. The probability of a monitor passing these checks is independent of other monitors. 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.

The expected value of the binomial distribution is:

E(X) = np

The variance of the binomial distribution is:

V(X) = np(1-p)

80% of the monitors made by your company will pass your stringent quality control checks.

This means that p = 0.8

40 monitors have just come off the assembly line.

This means that n = 40

a) What is the probability that exactly 30 of them will pass the quality control checks?

This is P(X = 30). So

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

P(X = 30) = C_{40,30}.(0.8)^{30}.(0.2)^{10} = 0.1075

10.75% probability that exactly 30 of them will pass the quality control checks.

b) What are the expected value and variance of the number of these monitors that will pass the quality control checks?

E(X) = np = 40*0.8 = 32

Expected value is 32

V(X) = np(1-p) = 40*0.8*0.2 = 6.4

Variance is 6.4

4 0
3 years ago
The design of a nomadic tribal rug is shown, where m&lt;AED=34° and m&lt;EAD=112°
Lady_Fox [76]

The required angle m∠ADB  of the nomadic tribal rug is 112°

Given, that in ΔAED and ΔBDC,

where, m∠AED = 34° and m∠EAD = 112°

To find m∠ADB

By using the angle sum property of a triangle, which states that the sum of interior angles of a triangle is 180°.

Therefore in ΔAED,

m∠AED + m∠EAD +  m∠ADE = 180°

   34°  +    112°    +   m∠ADE   = 180°

or,    m∠ADE + 146° = 180°

or,       m∠ADE    = 34°

Similarly, m∠BDC = 34°

Now, by using the sum of linear pair of angles,

m∠ADE + m∠ADB + m∠BDC = 180°

  34° + m∠ADB + 34°  = 180°

   m∠ADB + 68° = 180°

   m∠ADB  = 180°- 68°

    m∠ADB = 112°

To learn more about linear pair of angles, visit: brainly.com/question/26555759?referrer=searchResults

#SPJ9

5 0
2 years ago
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