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2 years ago

Length of the missing leg is:

Mathematics

1 answer:

ad-work [718]2 years ago

5 0

$formula = {a}^{2} + {b}^{2} = {c}^{2}$

Answers and work are on the attached sheet! :)

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Pick the correct description of the line 2y = 10

lana [24]

2y = 10
y = 10/2
y = 5 <==horizontal

this is a horizontal line. Anytime you have y = an integer, it is horizontal...and if u have x = an integer, it is a vertical line.

8 0

2 years ago

Read 2 more answers

3(2x) + 3(-1) = -21 6x - 3 = -21 What property is being used ?

kaheart [24]

Distributive property, since 3(2x) you distribute and get 6x

4 0

2 years ago

It is well-defined collection of distinct objects A.class B.elements C.Group D.Set

Elena L [17]

Answer

hope correct

D

4 0

2 years ago

3x4²-(-6)²+4-6x4 Need help!! Please!!

goblinko [34]

Assuming the x stands for multiplication... follow the ordering rules:

First let's take care of the exponents. 4²=16 and (-6)²=36. At the same time I will add some spacing between the terms:

3x16 - 36 + 4 - 6x4

Now do the multiplications:

48 - 36 + 4 - 24

The rest is just addition and subtraction

48 - 36 + 4 - 24 = -8

4 0

2 years ago

g A population is infected with a certain infectious disease. It is known that 95% of the population has not contracted the dise

trasher [3.6K]

Answer:

There is approximately 17% chance of a person not having a disease if he or she has tested positive.

Step-by-step explanation:

Denote the events as follows:

D = a person has contracted the disease.

+ = a person tests positive

- = a person tests negative

The information provided is:

$P(D^{c})=0.95\\P(+|D) = 0.98\\P(+|D^{c})=0.01$

Compute the missing probabilities as follows:

$P(D) = 1- P(D^{c})=1-0.95=0.05\\\\P(-|D)=1-P(+|D)=1-0.98=0.02\\\\P(-|D^{c})=1-P(+|D^{c})=1-0.01=0.99$

The Bayes' theorem states that the conditional probability of an event, say A provided that another event B has already occurred is:

$P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^{c})P(A^{c})}$

Compute the probability that a random selected person does not have the infection if he or she has tested positive as follows:

$P(D^{c}|+)=\frac{P(+|D^{c})P(D^{c})}{P(+|D^{c})P(D^{c})+P(+|D)P(D)}$

$=\frac{(0.01\times 0.95)}{(0.01\times 0.95)+(0.98\times 0.05)}\\\\=\frac{0.0095}{0.0095+0.0475}\\\\=0.1666667\\\\\approx 0.1667$

So, there is approximately 17% chance of a person not having a disease if he or she has tested positive.

As the false negative rate of the test is 1%, this probability is not unusual considering the huge number of test done.

7 0

2 years ago