Answer:
BC:BN=8:3
Step-by-step explanation:
ABCD is a trapezoid and there is a point m which belongs to AD such that AM:MD=3:5.Line "l" parallel to AB intersects the diagonal AC at p and BD at N.
Now, we know that the parallel lines divide the transversal into the segments with equal ratio, therefore, BN:NC=AM:MD
But, BC= BN+NC
Therefore, BC:BN=(BN+NC):BN
⇒BC:BN=(3+5):3
⇒BC:BN=8:3
Answer: 0.05
Step-by-step explanation:
Let M = Event of getting an A in Marketing class.
S = Event of getting an A in Spanish class,
i.e. P(M) = 0.80 , P(S) = 0.60 and P(M∩S)=0.45
Required probability = P(neither M nor S)
= P(M'∩S')
= P(M∪S)' [∵P(A'∩B')=P(A∪B)']
=1- P(M∪S) [∵P(A')=1-P(A)]
= 1- (P(M)+P(S)- P(M∩S)) [∵P(A∪B)=P(A)+P(B)-P(A∩B)]
= 1- (0.80+0.60-0.45)
= 1- 0.95
= 0.05
hence, the probability that Helen does not get an A in either class= 0.05
Answer:
a. 12
b. 7.200 and 2.683
Step-by-step explanation:
The computation is shown below:
Given that
P = 0.40 and n = 30
a)
The expected value of received e-mails is
= n × p
= 30 × 0.4
= 12
b)
The variance of emails received is
= n × p × (1 - p)
= 30 × 0.4 × 0.6
= 7.200
Now
The standard deviation of emails received is
= sqrt(variance)
= 2.683
We simply applied the above formula
A counter example is
f(x)=5x^5+2x^3+3x
g(x)=-5x^5-x^4+x^2-4
Then f(x)+g(x) = -x^4+2x^3+x^2+3x-4 which is a polynomial of degree 4.
So the answer is no. Counter-example is given above.
Circumference<span> = 2 x </span>Radius<span> x Pi C = 2 * r * π, where: r = </span>radius<span> and π = </span>3.14<span> 1) All you have to do is substitute in order to find the </span>radius<span> (r) of the </span>bike tire<span>, C = 2πr 82.896 = 2 * </span>3.14<span> * r 82.896 = 6.28 * r r = 82.896 / 6.28 r = 13.2. Thus, r = 13.2 which means that the </span>radius<span> of the circle (</span>bike tire<span>) is 13.2 </span>inches<span> </span>
