All categories

3 years ago

Help plzzzzzzzzzzzz help me

Mathematics

1 answer:

r-ruslan [8.4K]3 years ago

8 0

Answer:

wow-

Step-by-step explanation:

ngl, thats hard.

You might be interested in

Factor by grouping:

erma4kov [3.2K]

The idea is to group up the terms into two groups, factor each group and then factor out the overall GCF

2x - 18 + xy - 9y
(2x - 18) + (xy - 9y)
2(x - 9) + (xy - 9y)
2(x - 9) + y(x - 9)
(2 + y)(x - 9)
(x - 9)(2 + y)

Answer: Choice B

3 0

4 years ago

Can pls someone help and show me how to solve i’m<br> really struggling

Nadya [2.5K]

Answer:

∠RQS = 156

Step-by-step explanation:

m∠TQS = 24

∠TQS and ∠RQS are linear pair. Linear pair are two adjacent angles and their sum is 180

∠TQS + ∠RQS = 180

24 + ∠RQS = 180

∠RQS = 180 - 24

∠RQS = 156

5 0

3 years ago

Read 2 more answers

#1: Consider the list of golf scores: 71, 68, 72, 79, 80, 77, 84, 70, 92. What is

Nesterboy [21]

Answer:none

Step-by-step explanation:

8 0

3 years ago

A number is three less than six times another numbers. The sum is 53. Find the numbers

Pachacha [2.7K]

X = 6y - 3
x + y = 53

6y - 3 + y = 53
7y - 3 = 53
7y = 53 + 3
7y = 56
y = 56/7
y = 8

x = 6y - 3
x = 6(8) - 3
x = 48 - 3
x = 45

ur numbers are : 8 and 45

3 0

3 years ago

Suppose that one person in 10,000 people has a rare genetic disease. There is an excellent test for the disease; 98.8% of the pe

nirvana33 [79]

Answer:

A)The probability that someone who tests positive has the disease is 0.9995

B)The probability that someone who tests negative does not have the disease is 0.99999

Step-by-step explanation:

Let D be the event that a person has a disease

Let $D^c$ be the event that a person don't have a disease

Let A be the event that a person is tested positive for that disease.

P(D|A) = Probability that someone has a disease given that he tests positive.

We are given that There is an excellent test for the disease; 98.8% of the people with the disease test positive

So, P(A|D)=probability that a person is tested positive given he has a disease = 0.988

We are also given that one person in 10,000 people has a rare genetic disease.

So, $P(D)=\frac{1}{10000}$

Only 0.4% of the people who don't have it test positive.

$P(A|D^c)$ = probability that a person is tested positive given he don't have a disease = 0.004

$P(D^c)=1-\frac{1}{10000}$

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

$P(D|A)=\frac{0.988 \times \frac{1}{10000}}{0.988 \times (1-\frac{1}{10000}))+0.004 \times (1-\frac{1}{10000})}$

P(D|A)= $\frac{2470}{2471}$ =0.9995

P(D|A)= $0.9995$

A)The probability that someone who tests positive has the disease is 0.9995

(B)

$P(D^c|A^c)$ =probability that someone does not have disease given that he tests negative

$P(A^c|D^c)$ =probability that a person tests negative given that he does not have disease =1-0.004

=0.996

$P(A^c|D)$ =probability that a person tests negative given that he has a disease =1-0.988=0.012

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

$P(D^c|A^c)=\frac{0.996 \times (1-\frac{1}{10000})}{0.996 \times (1-\frac{1}{10000})+0.012 \times \frac{1}{1000}}$

$P(D^c|A^c)=0.99999$

B)The probability that someone who tests negative does not have the disease is 0.99999

8 0

4 years ago