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3 years ago

What is 1/3. of 4/5 I need help

Mathematics

2 answers:

levacccp [35]3 years ago

6 0

To do this multiply 1 by 4, then 3 by 5. 4 over 15 is your answer.

Sauron [17]3 years ago

4 0

Answer:

4/15

Step-by-step explanation:

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Pleaseee helpp!!!!!!

ludmilkaskok [199]

Answer:

Step-by-step explanation:

because she uses beads

so the number of beads go down

6 0

4 years ago

Read 2 more answers

Only need help on number 5 and 6

Alborosie

5) since (x-a)^2+(y-b)^2=r^2

plug in (-5,2) as (a,b) and r=3

Making the equation:

(x+5)^2+(y-2)^2=9

6) (x-0)^2+(y-0)^2=r^2

since we have (0,5)

0^2+5^2=r^2

Final equation: x^2+y^2=25

5 0

3 years ago

Help me -2 (8 - e) = 2 - e

-BARSIC- [3]

Answer:

e = 6

Step-by-step explanation:

Step 1: Write equation

-2(8 - e) = 2 - e

Step 2: Solve for <em>e</em>

Distribute -2: -16 + 2e = 2 - e
Add e to both sides: -16 + 3e = 2
Add 16 to both sides: 3e = 18
Divide both sides by 3: e = 6

Step 3: Check

<em>Plug in e to verify it's a solution.</em>

-2(8 - 6) = 2 - 6

-2(2) = -4

-4 = -4

3 0

3 years ago

What is the average of 1/4 and 6/18

galina1969 [7]

Answer:

7/24

Step-by-step explanation:

(1/4+6/18)/2=7/24

8 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

$P(A\cupB) = P(A)+P(B)-P(A\cap B) = P(A) +P(B)-P(A|B)P(B) = 6.4\%+5\%-90\%\cdot 5\%=6.9\%$

4 0

4 years ago