To find the total loss for this play, take the two values, find their signs (positive or negative), then sum them. In this case, the play itself resulted in a movement of -12.5 yards, and the penalty added a -5 yard movement to the result. This would lead to a total movement of (-12.5 + [-5]), or -17.5 yards.

4 0

3 years ago

Find the exact value of tan 15 degrees

pentagon [3]

tan(15) =
sin(15) / cos(15) =
sin(45 - 30) / cos(45 - 30) =
[ sin(45)cos(30) - sin(30)cos(45) ] / [ cos(45)cos(30) + sin(45)sin(30)]

Since sin(45) = cos(45) = √2/2, you can just factor that out from the top and bottom
[ cos(30) - sin(30) ] / [ cos(30) + sin(30)]
[ √3/2 - 1/2 ] / [ √3/2 + 1/2]
(√3 - 1) / (√3 + 1)
(√3 - 1)^2 / (√3+1)(√3 - 1)
(√3 - 1)^2 / (3 - 1)
(3 - 2√3 +1) / 2
2 - √3

There's also a formula for tan(a-b), but I couldn't remember it off hand.

4 0

3 years ago

A couple intends to have two children, and suppose that approximately 52% of births are male and 48% are female.

Pachacha [2.7K]

a) Probability of both being males is 27%

b) Probability of both being females is 23%

c) Probability of having exactly one male and one female is 50%

Step-by-step explanation:

The probability that the birth is a male can be written as

$p(m) = 0.52$ (which corresponds to 52%)

While the probability that the birth is a female can be written as

$p(f) = 0.48$ (which corresponds to 48%)

Here we want to calculate the probability that over 2 births, both are male. Since the two births are two independent events (the probability of the 2nd to be a male does not depend on the fact that the 1st one is a male), then the probability of both being males is given by the product of the individual probabilities:

$p(mm)=p(m)\cdot p(m)$

And substituting, we find

$p(mm)=0.52\cdot 0.52 = 0.27$

So, 27%.

In this case, we want to find the probability that both children are female, so the probability

$p(ff)$

As in the previous case, the probability of the 2nd child to be a female is independent from whether the 1st one is a male or a female: therefore, we can apply the rule for independent events, and this means that the probability that both children are females is the product of the individual probability of a child being a female:

$p(ff)=p(f)\cdot p(f)$