4 plus radical 3 times 4 plus radical 3

2 Points 8 is a factor of 24. A. True B. False

Brums [2.3K]

Answer: true

Step-by-step explanation:

8 multiplied by 3 is 24

3 0

3 years ago

Read 2 more answers

10+(4x(13-5))-6 order of operations show ur work

Jobisdone [24]

I’m pretty sure your answer would be 36

3 0

3 years ago

Please help for a brainlist ;))

Nuetrik [128]

The answer is option A.

The $50 initial membership fee represents the y-intercept (b), while the $20 monthly membership fee represents the slope (m) of the equation.

6 0

3 years ago

What is the equation of the axis of symmetry of the graph of y=ax^2+bx+c

Phantasy [73]

Answer:

$x=-\frac{b}{2a}$

Step-by-step explanation:

we know that

The quadratic equation $y=ax^{2}+bx+c$ is a vertical parabola

The axis of symmetry in a vertical parabola is a vertical line

The equation of the axis of symmetry in a vertical parabola is equal to the x-coordinate of the vertex

The x-coordinate of the vertex is equal to the formula

$x=-\frac{b}{2a}$

therefore

the equation of the axis of symmetry is equal to $x=-\frac{b}{2a}$

4 0

3 years ago

Read 2 more answers

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