What is the 7th term in the geometric sequence described by this explicit formula?

Answer C? Reflexive SSS?

BARSIC [14]

Yes the answer is c. C is correct

8 0

4 years ago

The diameter of a bicycle wheel is 63 cm. How much distance will it cover in 20revolutions?

alina1380 [7]

Answer:

1256 cm

Step-by-step explanation:

First we must find the perimeter, because the perimeter is equal to 1 revolution.

The perimeter of any circle is Pi(Diameter)

3.14*20

62.8 cm is the perimeter of the bike wheel.

Because the perimeter is basically a revolution, we can multiply 62.8 by 20 to get:

1256 cm

7 0

2 years ago

Evaluate g(4): g(x) = 8x2 + 9x – 7 A) 121 B) 135 C) 154 D) 157

Lunna [17]

Answer:

g(4) = 157

Step-by-step explanation:

Substitute 4 for x in g(x) = 8x^2 + 9x – 7, obtaining:

g(4) = 8(4)^2 + 9(4) - 7 = 128 + 36 - 7 = 157

8 0

3 years ago

Give the numerical value of the parameter p in the following binomial distribution scenarioA softball pitcher has a 0.721 probab

Igoryamba

Answer:

$P(X>15)= P(X=16)+P(X=17)+P(X=18)+P(X=19)$

$P(X=16)=(19C16)(0.721)^{16} (1-0.721)^{19-16}=0.112$

$P(X=17)=(19C17)(0.721)^{17} (1-0.721)^{19-17}=0.051$

$P(X=18)=(19C18)(0.721)^{18} (1-0.721)^{19-18}=0.015$

$P(X=19)=(19C19)(0.721)^{19} (1-0.721)^{19-19}=0.002$

And replacing we got:

$P(X>15)= P(X=16)+P(X=17)+P(X=18)+P(X=19) =0.112+0.051+0.015+0.002= 0.1801$

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Solution to the problem

For this case our random variable is given by:

$X \sim Binom(n = 19, p = 0.721)$

For this case we want this probability:

$P(X>15)= P(X=16)+P(X=17)+P(X=18)+P(X=19)$

$P(X=16)=(19C16)(0.721)^{16} (1-0.721)^{19-16}=0.112$

$P(X=17)=(19C17)(0.721)^{17} (1-0.721)^{19-17}=0.051$

$P(X=18)=(19C18)(0.721)^{18} (1-0.721)^{19-18}=0.015$

$P(X=19)=(19C19)(0.721)^{19} (1-0.721)^{19-19}=0.002$

And replacing we got:

$P(X>15)= P(X=16)+P(X=17)+P(X=18)+P(X=19) =0.112+0.051+0.015+0.002= 0.1801$

$P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668$

3 0

4 years ago

Click on icon to see problem, By hand(no calculator)

anastassius [24]

Let t=reference angle of theta=asin(3/5) in Q1
sin(theta)=3/5 in Q2 =>
cos(theta)=cos(pi-t)=-cos(t)=-4/5
(a) sin(pi/6)=sin(30)=1/2, cos(pi/6)=cos(30)=sqrt(3)/2
sin(theta+pi/6)
=sin(theta)cos(pi/6)+cos(theta)sin(pi/6)
=(3/5)(sqrt(3)/2)+(-4/5)(1/2)
=(3sqrt(3)-4)/10 [ after a few simplifications]

(b)
cos(5pi/3)=cos(pi+2pi/3)=cos(pi)cos(2pi/3)-sin(pi)sin(2pi/3)=-1(-1/2)-0=1/2
sin(5pi/3)=sin(pi+2pi/3)=sin(pi)cos(2pi/3)+cos(pi)sin(2pi/3)=0+(-1)(sqrt(3)/2)=-sqrt(3)/2

cos(5pi/3-theta)
=cos(5pi/3)cos(theta)+sin(5pi/3)sin(theta)
=(1/2)(-4/5)+(-sqrt(3)/2)(3/5)
=-(3sqrt(3)+4)/10

(c)
cos(2theta)
=cos^2(theta)-sin^2(theta)
=(-4/5)^2-(3/5)^2
=7/25

(d)
csc(pi/2-theta)
=1/sin(pi/2-theta)
=1/(sin(pi/2)cos(theta)-cos(pi/2)sin(theta))
=1/[(1)(-4/5)-0(3/5)]
=1/[-4/5]
=-5/4

4 0

3 years ago