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3 years ago

You have $6.17. you buy 7 items that cost $0.76 each. how much money do you have left

Mathematics

2 answers:

Shkiper50 [21]3 years ago

6 0

Answer:

$0.85

Step-by-step explanation:

if you multiply 7x0.76=5.32

and then subtract the 6.17-5.32=$.85

Klio2033 [76]3 years ago

5 0

Answer:

.85

Step-by-step explanation:

Multiply .76x7 then subtract 5.32 from 6.17 and you get 0.85.

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2y = 4

inessss [21]

C) (3, 2)
Solve for x and y, then plug in (x, y)
2y = 4, 2y/2 = 4/2, y = 2
-x = -3, -x/-1 = -3/-1, x = 3

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3 years ago

You don't have to do all of them

vova2212 [387]

Answer: lol your using canvas

1. 63

2. D

3. B

4. A

Sorry for the bland explinations havin a bad day.

Step-by-step explanation:

1. The way we solve this is by adding the degrees 72 and 45. then we subtract it by 180. this is because in all triangles the angles always add up to 180 degrees.

2. Like question one what set of angles measures to 180 degrees.

3. The sides of a triangle should define the angle measurements. you can also see if its a triangle if the numbers are continues like 3,4,5 or in this case 4,5,6

4. like in questions 1 and 2 the angles of a triangle has to equal 180 degrees so 60 times 3 is 180 so each angle would have to be 60 degrees.

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3 years ago

The graph of the following system of equations is

eimsori [14]

The answer is <span>C. Intersecting lines</span>

3 0

3 years ago

Read 2 more answers

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

Which expression is equivalent to -<br>60x20, 24<br>30x10 12?

hoa [83]

Answer:

30x10 12

Step-by-step explanation:

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4 years ago