How solutions are in 25 - 4x = 15 - 3x + 10

sashaice [31]

Use b•h\2 to solve the triangular prisms rectangular use l•w•h

8 0

3 years ago

Ethan Doran is a physical therapist. His biweekly salary is $3,985. What is his annual salary? (Hint: You need to know how many

NNADVOKAT [17]

103,610.00 you welcome.

7 0

3 years ago

If p is true and ~ q is false, then p ~ q is always, sometimes, never false.

Paha777 [63]

Answer:

never

Step-by-step explanation:

The expression: ~q means "not q"

If ~q is false, then q is true

The expression:

If p is true and ~q is false

is equivalent to:

If p is true and q is true

then:

If p is true and q is true, then p -> q is never false (or is always true, see table attached). Or:

If p is true and ~q is false, then p -> ~q is never false.

7 0

3 years ago

A survey conducted by the Consumer Reports National Research Center reported, among other things, that women spend an average of

Nookie1986 [14]

Answer:

(a) The probability that a randomly selected woman shop exactly two hours online is 0.217.

(b) The probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c) The probability that a randomly selected woman shop less than 5 hours online is 0.9922.

Step-by-step explanation:

Let X = time spent per week shopping online.

It is provided that the random variable X follows a Poisson distribution.

The probability function of a Poisson distribution is:

$P (X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0,1,2,...$

The average time spent per week shopping online is, λ = 1.2.

(a)

Compute the probability that a randomly selected woman shop exactly two hours online over a one-week period as follows:

$P (X=2)=\frac{e^{1.2}(1.2)^{2}}{2!} =0.21686\approx0.217$

Thus, the probability that a randomly selected woman shop exactly two hours online is 0.217.

(b)

Compute the probability that a randomly selected woman shop 4 or more hours online over a one-week period as follows:

P (X ≥ 4) = 1 - P (X < 4)

= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

$=1-\frac{e^{1.2}(1.2)^{0}}{0!}-\frac{e^{1.2}(1.2)^{1}}{1!}-\frac{e^{1.2}(1.2)^{2}}{2!}-\frac{e^{1.2}(1.2)^{2}}{3!}\\=1-0.3012-0.3614-0.2169-0.0867\\=0.0338$

Thus, the probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c)

Compute the probability that a randomly selected woman shop less than 5 hours online over a one-week period as follows:

P (X < 5) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

$=\frac{e^{1.2}(1.2)^{0}}{0!}+\frac{e^{1.2}(1.2)^{1}}{1!}+\frac{e^{1.2}(1.2)^{2}}{2!}+\frac{e^{1.2}(1.2)^{3}}{3!}+\frac{e^{1.2}(1.2)^{4}}{4!}\\=0.3012+0.3614+0.2169+0.0867+0.0260\\=0.9922$

Thus, the probability that a randomly selected woman shop less than 5 hours online is 0.9922.

8 0

4 years ago

Write the slope-intercept form of the equation of the line that is perpendicular to AB and passes through Point X Show

Fynjy0 [20]

Answer:

y = 12/5x + 22

Step-by-step explanation:

First, let's find the slope of the AB line.

y2 - y1 / x2 - x1

3 - 8 / 2 - (-10)

-5 / 12

= -5/12

When a line is perpendicular to something, the slope becomes the opposite.

y = 12/5x + b

10 = 12/5(-5) + b

10 = -12 + b

22 = b

4 0

3 years ago

How solutions are in 25 - 4x = 15 - 3x + 10 - x