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cluponka [151]
3 years ago
7

Please answer soooooooooooooooon :|

Mathematics
1 answer:
madam [21]3 years ago
4 0

Answer: 200 square units

Step-by-step explanation:

first you think what 60/3 is and you get 20

then you multiply 20x20 and get 400

then you divide by 2 because to find the area of a triangle you use the formula (lxh)/2

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I need help with this ?
rewona [7]

Answer:

Range: [-5, 0]

General Formulas and Concepts:

<u>Algebra I</u>

  • Range is the set of y-values that are outputted by function f(x)

Step-by-step explanation:

According to the graph, the line's y-values span from -5 to 0. Since both are closed dot, they are inclusive in the range:

[-5, 0] or -5 ≤ y ≤ 0

5 0
3 years ago
Find the surface area of the cylinder 6cm 6cm <br> Give your answer in terms of pi
Lisa [10]

Answer:  A=2πrh+2πr2

Hope this helps you out! :}


4 0
3 years ago
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"Woohoo! The school carnival is here! You think about all the fun carnival games that you love to play every year. Then you reme
Lena [83]

Answer:

14

Step-by-step explanation:

5 0
3 years ago
Solve for x: 3x-10 = 2(x+6)
77julia77 [94]

Answer:

x = 22

Step-by-step explanation:

7 0
3 years ago
A diagnostic test for a disease is such that it (correctly) detects the disease in 90% of the individuals who actually have the
Ne4ueva [31]

Answer:

0.2177 = 21.77% conditional probability that she does, in fact, have the disease

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Test positive

Event B: Has the disease

Probability of a positive test:

90% of 3%(has the disease).

1 - 0.9 = 0.1 = 10% of 97%(does not have the disease). So

P(A) = 0.90*0.03 + 0.1*0.97 = 0.124

Intersection of A and B:

Positive test and has the disease, so 90% of 3%

P(A \cap B) = 0.9*0.03 = 0.027

What is the conditional probability that she does, in fact, have the disease

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.027}{0.124} = 0.2177

0.2177 = 21.77% conditional probability that she does, in fact, have the disease

3 0
3 years ago
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