Simplify the problem 7x+4(6-x)+28

What are the zeros of this function? look at graph a. b. c. d.

fredd [130]

The zeros of the function occur when the graph meet the x-axis so in this case the zeros occur at B ( 0 and 5)

4 0

2 years ago

Evaluate the expression 10b/a if b = 6 and a = 2

zimovet [89]

Answer:

30

Step-by-step explanation:

10b/a

b = 6

a =2

10(6)/2

10x6 / 2 = 60/2

= 30

8 0

2 years ago

Read 2 more answers

Help with this math question please

Cloud [144]

A = P( 1 + rt)
A/P = 1 + rt
A/P - 1 = rt
t = (( A/P)) - 1)/r

t = (( 5500/1000) - 1)/(6.25/100
t = (5.5 - 1)/0.0625
t = 4.5/0.0625 = 72 years

Answer it will take 72 Years...

Hope it helps!!!!!

4 0

3 years ago

Read 2 more answers

2. What is the interpretation of the constant of proportionality?

Snowcat [4.5K]

Answer: It would be the first one, 0.001 kilometers per meter.

Step-by-step explanation: is 1,000 meters is equal to 1 kilometer, you would have to move the decimal on each until one is at one and the other is at 0.001. This would leave you with the first answer. Hope this helps!

6 0

2 years ago

Suppose that X has an exponential distribution with mean equal to 10. Determine the following: a. P(X > 10) b. P(X > 20) c

GrogVix [38]

Answer:

(a) The value of P (X > 10) is 0.3679.

(b) The value of P (X > 20) is 0.1353.

(c) The value of P (X < 30) is 0.9502.

(d) The value of x is 30.

Step-by-step explanation:

The probability density function of an exponential distribution is:

$f(x)=\lambda e^{-\lambda x};\ x>0, \lambda>0$

The value of E (X) is 10.

The parameter λ is:

$\lambda=\frac{1}{E(X)}=\frac{1}{10}=0.10$

(a)

Compute the value of P (X > 10) as follows:

$P(X>10)=\int\limits^{\infty}_{10} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{10} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{10}\\=|e^{-0.10 x} |^{\infty}_{10}\\=e^{-0.10\times10}\\=0.3679$

Thus, the value of P (X > 10) is 0.3679.

(b)

Compute the value of P (X > 20) as follows:

$P(X>20)=\int\limits^{\infty}_{20} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{20} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{20}\\=|e^{-0.10 x} |^{\infty}_{20}\\=e^{-0.10\times20}\\=0.1353$