The point (2,-5) is reflected over the y-axis. Which of the following points is the image?

What is the minimum number of fish in the pond? (Hint: You caught 100 fish both days, but you caught 3 of the same fish twice.)

nevsk [136]

Answer:

197

Step-by-step explanation:

So, if you catch 100 fish on both days, that means that you have caught 200 fish. 100 on one day and 100 on the next. But, since you've caught 3 fish twice, that means that three out of 200 fish are the same and cannot be counted as new fish. So 3 from 200 is 197 fish. Hope this helped!

Day 1: 100 (3 tagged)

Day 2: 97 (3 tagged)

100+97=197

or

Day 1: 100 (3 tagged)

Day 2: 100 (3 tagged)

100+100= 200

200-3=197

Answer, 197 fish.

8 0

2 years ago

Find the 30th term of an arithmetic sequence whose first term is 2 and common difference is 5.

Veronika [31]

2 + 29 times 5 = 147

3 0

2 years ago

Read 2 more answers

How do I calculate the Pythagorean theorem?

kogti [31]

A^2 + b^2 = c^2
A*2 b*2
When you get the final answer of a and b after squaring you then add a and b and you get your answer C^2

7 0

3 years ago

Yooo plz help asap!!! marking brainiest answer!!!

Komok [63]

Answer:

C.

Step-by-step explanation:

You are looking for an equation that, when you input 0 for x, you get 4 for y, and when you input -5 for x, you get -1 for y.

That only works for Option C.

Hope this [sorta] helps!

6 0

2 years ago

Read 2 more answers

certain virus infects one in every 300 people. A test used to detect the virus in a person is positive 80% of the time if the pe

Alekssandra [29.7K]

Answer:

The probability that a person has the virus given that they have tested positive is P(A|B)=0.032.

The probability that a person does not have the virus given that they have tested negative is P(A'|B')=0.999.

Step-by-step explanation:

a) The probability of B: "the person tests positive" is the sum of:

- The probability that a person has the virus and the test is positive.

$P(A) \cdot P(B|A)$

- The probability that a person don't have the virus but the test is positive.

$P(not \,A) \cdot P(B|not\,A)$

Then, the probability of B: "the person tests positive" is

$P(A|B)=\frac{P(B|A)}{P(B)}=\frac{P(A)\cdot P(B|A)}{P(A)\cdot P(B|A)+P(A')\cdot P(B|A')}$

We have

$P(A)=1/300=0.0033\\\\P(A')=1-P(A)=0.9967\\\\P(B|A)=0.8\\\\P(B|A')=0.08$

Then

$P(A|B)=\frac{P(A)\cdot P(B|A)}{P(A)\cdot P(B|A)+P(A')\cdot P(B|A')}\\\\P(A|B)=\frac{0.0033*0.8}{0.0033*0.8+0.9967*0.08}= \frac{0.00264}{0.00264+0.07974}=\frac{0.00264}{0.08238}= 0.032$

The probability that a person has the virus given that they have tested positive is P(A|B)=0.032.

b) The probability P(A'|B') is

$P(A'|B')=\frac{P(B'|A')\cdot P(A')}{P(B'|A')\cdot P(A')+P(B'|A)\cdot P(A)}$

We have

$P(A)=0.0033\\\\P(A')=1-P(A)=0.9967\\\\P(B'|A)=1-0.8=0.2\\\\P(B'|A')=1-0.08=0.92$

Then

$P(A'|B')=\frac{P(B'|A')\cdot P(A')}{P(B'|A')\cdot P(A')+P(B'|A)\cdot P(A)}=\frac{0.92*0.9967}{0.92*0.9967+0.2*0.0033}= \frac{0.917}{0.917+0.00066}= 0.999$

The probability that a person does not have the virus given that they have tested negative is P(A'|B')=0.999.

5 0

3 years ago