2/3 x -9/8 x -4/5 x -1

Mathematics

1 answer:

jolli1 [7]2 years ago

7 0

-151/120x -1 or 1\120 × (-151x -120)

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A sequence is defined by the function F(n) = F(n -1) +5, where n represents the number of the term for n >1, and f(1) = -4. W

Anit [1.1K]

Answer:

4th option

Step-by-step explanation:

Using the recursive rule and f(1) = - 4 , then

f(2) = f(1) + 5 = - 4 + 5 = 1

f(3) = f(2) + 5 = 1 + 5 = 6

f(4) = f(3) + 5 = 6 + 5 = 11

The first 4 terms are - 4, 1, 6, 11

5 0

2 years ago

I need help ASAP on Tangent Circles!!

loris [4]

I will solve one for you to get you a idea so next u can solve all
To know whether line is tangent toa circle or not we should know that that line is a tangent line of a circle if it intersects circle at single point.
and this will always be perpendicular at radius
so in problem
1)

diameter is given 7.5
and u can see they make a right triangle ABC if that is perpendicular so segment AB is tangent to a circle.
Lets see by applying <span>Pythagorean Theorem
</span>this is method we can say that this is in fact a right triangle

a^2=b^2=c^2
a,b are shorter side c is longer side
(7.5)^2+(8)^2=(17)^2
56.25+64=289
120.25=289
both are not equal so they are not forming right triangle so segment AB is not a tangent line.

5 0

2 years ago

A car dealership sells 0, 1, or 2 luxury cars on any day. When selling a car, the dealer also tries to persuade the customer to

melisa1 [442]

Answer:

Mean = 1.42

Variance = 0.58

Step-by-step explanation:

Given: X denote the number of luxury cars sold in a given day, and Y denote the number of extended warranties sold.

Also, joint probability function of X and Y are given.

To find:

mean and variance of X

Solution:

From the given joint probability function of X and Y,

$P(X=0)=\frac{1}{6}\\P(X=1)=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}=\frac{3}{12}\\P(X=2)=\frac{1}{12}+\frac{1}{3}+\frac{1}{6}=\frac{1+4+2}{12}=\frac{7}{12}$

Mean of X:

$E(X)=\sum XP(X)\\=0\left ( \frac{1}{6} \right )+1\left ( \frac{3}{12} \right )+2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{14}{12}\\=\frac{17}{12}=1.42$

Variance of X:

$E(X^2)=\sum X^2P(X)\\=0^2\left ( \frac{1}{6} \right )+1^2\left ( \frac{3}{12} \right )+2^2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{28}{12}\\=\frac{31}{12}$

$var(X)=E\left [ X^2 \right ]-\left ( E\left [ X \right ] \right )^2\\=\frac{31}{12}-\left ( \frac{17}{12} \right )^2\\=\frac{31}{12}-\frac{289}{144}\\=\frac{372-289}{144}\\=\frac{83}{144}\\=0.58$