What is the standard equation with center (4, -5) and radius 11

A soccer player runs a total of 12 miles during a 4 hour game.<br> What is the unit rate?

VARVARA [1.3K]

Answer:

3

Step-by-step explanation:

12 divited by 4 =3

6 0

3 years ago

Meijer sells 10 bottles of soap for $15 and 20 bottles of soap for

podryga [215]

Answer:

20 bottles for 28

Step-by-step explanation:

15 ÷ 10 = 1.5

28 ÷ 20 = 1.4

3 0

3 years ago

(1 point) Carbon dating is often used to determine the age of a fossil. For example, a humanoid skull was found in a cave in Sou

Aleksandr [31]

Answer:

The estimated age of the skull is 39118 years.

Step-by-step explanation:

The amount of the substance after t years is given by:

$A(t) = A(0)(1-r)^t$

In which A(0) is the initial amount, and r is the decay rate.

The half-life of carbon-14 is about 5600 years.

This means that $A(5600) = 0.5A(0)$ . We use this to find r, or 1 - r, to replace in the equation. So

$A(t) = A(0)(1-r)^t$

$0.5A(0) = A(0)(1-r)^{5600}$

$(1-r)^{5600} = 0.5$

$\sqrt[5600]{(1-r)^{5600}} = \sqrt[5600]{0.5}$

$1 - r = (0.5)^{\frac{1}{5600}}$

$1 - r = 0.9999$

So

$A(t) = A(0)(0.9999)^t$

Only 2% of the original amount of carbon-14 remains in the burnt wood of the campfire.

This is t for which $A(t) = 0.02A(0)$ . So

$A(t) = A(0)(0.9999)^t$

$0.02A(0) = A(0)(0.9999)^t$

$(0.9999)^t = 0.02$

$\log{(0.9999)^t} = \log{0.02}$

$t\log{0.9999} = \log{0.02}$

$t = \frac{\log{0.02}}{\log{0.9999}}$

$t = 39118$

The estimated age of the skull is 39118 years.

6 0

3 years ago

Will give brainliest if correct.

suter [353]

┐( ∵ )┌

I think it's 4x? I'm not up to that level of math yet.

6 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