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2 years ago

In which quadrant the values of x and y both negative.

Mathematics

1 answer:

HACTEHA [7]2 years ago

4 0

Answer:

The Answer should be Quadrant III

Step-by-step explanation:

Hope this helped!

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Please need help on this one

tekilochka [14]

On which one do you need help with?

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3 years ago

Help me plz its algebra 1

kvv77 [185]

Answer:

C. x > -7 and x <= 3

Step-by-step explanation:

-3 < x + 4 <= 7

Subtract 4 from the three expressions.

-7 < x <= 3

-7 < x means x > -7.

Then you have x <= 3.

Answer: C. x > -7 and x <= 3

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3 years ago

9.<br> The cost of 20 litres of petrol is £18<br> Work out the cost of 1 litre of petrol.

yulyashka [42]

Answer:

20 x 1 / 18= 1.11

Step-by-step explanation:

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2 years ago

Read 2 more answers

What is the difference between -93 and 114 on a number line?

zhenek [66]

Distance: 7-(-3)=10 Midpoint: (7+(-3))=4/2=2 hahaha haha thats right

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2 years ago

Refer to the Trowbridge Manufacturing example in Problem 2-35. The quality control inspection proce- dure is to select 6 items,

Ivanshal [37]

Answer:

77.64% probability that there will be 0 or 1 defects in a sample of 6.

Step-by-step explanation:

For each item, there are only two possible outcomes. Either it is defective, or it is not. The probability of an item being defective is independent of other items. 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 true proportion of defects is 0.15

This means that $p = 0.15$

Sample of 6:

This means that $n = 6$

What is the probability that there will be 0 or 1 defects in a sample of 6?

$P(X \leq 1) = P(X = 0) + P(X = 1)$

In which

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

$P(X = 0) = C_{6,0}.(0.15)^{0}.(0.85)^{6} = 0.3771$

$P(X = 1) = C_{6,1}.(0.15)^{1}.(0.85)^{5} = 0.3993$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.3771 + 0.3993 = 0.7764$

77.64% probability that there will be 0 or 1 defects in a sample of 6.

5 0

2 years ago