All categories

3 years ago

Plz help so i can talk to friends

Mathematics

2 answers:

dimulka [17.4K]3 years ago

5 0

I THINK IF I READ THAT RIGHT XD

9..IS A

10...IS D

viktelen [127]3 years ago

4 0

9.
A quadrilateral has 4 vertices. If you add two bases, its like adding two vertices. 4+2=6. A figure that has 6 vertices is a hexagon, so the answer is hexagon.

10.
Rhombus

Good luck!

You might be interested in

2 1/2 = 5n<br><br><br>use the inverse operation and show steps!!

blagie [28]

5 (n/2+1/2+1/2),1/2+5n+1

4 0

3 years ago

Ella has a box of chocolate candies. She gives 1/3 of the candies to her sister, 4 to her brother, and eats the remaining 12 can

tatyana61 [14]

Answer:

24 candies originally

Step-by-step explanation:

Add the 4 to how many you had left, then think about that number, 16, and what goes into 16 evenly 2 times? 8 and originally she gave away 1/3 of the box and if you were to add another 8 to the box it would be 24 divided by 1/3, -,-4,= 12.

3 0

3 years ago

All athletes at the Olympic Games (OG) are tested for performance-enhancing steroid drug use. The basic Anabolic Steroid Test (A

Semenov [28]

Answer:

a ) the probability of using steroids and having a negative test is 0.5%

b) The probability of testing positive is 6.4%

c) The probability of not using steroids, given that the test is negative is 99.47%

d) No, they are not statistically indepent.

e) The probability that the athlete will either use steroids or test positive is 6.9%

Step-by-step explanation:

Let A be the event that the test result is positive and B the event that the athlete uses Steroids. We are given the following

$P(A|B) = 90%, P(A|B^c) = 2%, P(B) = 5%$

From which we deduce that

$P(A^c|B) = 10%, P(B^c) = 95%$

a) We are asked for the probability $P(A^c\cap B)$ . REcall the conditional probability formula that, given two events C,D the conditional probability $P(C|D) = \frac{P(C\cap D)}{P(D)}$ . Then we have that

$P(A^c\cap B) = P(A^c|B)P(B) = 10\% \cdot 5\%=0.5\%$ .

b) We are asked for the probability P(A). We can use the fact that given two mutually exclusive events(that is, whose intersection is empty) A,B the probability P(C) of an event is given by $P(C) = P(C|A)P(A)+P(C|B)P(B)$ . Then

$P(A) = P(A|B)P(B)+P(A|B^c)P(B^c) = 90\%\cdot5\% + 2\% \cdot 95% = 6.4\%$

c) We are asked for the probability P(B^c|A^c). Recall that $P(A|B) = \frac{P(B|A)P(A)}{P(B)}$ . Then

$P(B^c|A^c) = \frac{P(A^c|B^c)P(B^c)}{P(A^c)}= \frac{P(A^c|B^c)P(B^c)}{1-P(A)}= \frac{(1-P(A|B^c))P(B^c)}{1-P(A)}=\frac{98\%\cdot 95\%}{1-6.4\%}= 99.47\%$

d) We say that two events A,B are statistically indepent if P(A|B) = P(A). Note that from point B the probability of testing negative is 1- 6.4% = 93.6%. Since 93.6% is different from 99.47% this means that testing positive and using steroids are not statistically independent.

e) We are asked for the probability $P(A\cup B)$ . We use the following