The quotient of x and 2 is greater than −30

For each function below, identify and enter the percent rate of change per unit, t.

exis [7]

Answer:

My answer isnt right when I checked. Sorry I couldn't help. :(

Step-by-step explanation:

3 0

4 years ago

Find the area of this regular polygon round to nearest tenth

Bezzdna [24]

Answer:

$A=163.9in^2$

Step-by-step explanation:

Let's use the regular polygon area formula when given the radius.

$A=\frac{(r^{2})(n)sin(\frac{360}{n})}{2}$

r is radius

n is the number of sides

Let's plug in the proper values into this equation and simplify:

$A=\frac{(7.74^{2})(7)sin(\frac{360}{7})}{2}$

$A=\frac{(59.9076)(7)(0.78183148)}{2}$

$A=\frac{327.86353299}{2}$

$A=163.93176649$

4 0

4 years ago

The number of years a person that age is expected to have left to live on

cupoosta [38]

Answer: I think its A

3 0

3 years ago

Read 2 more answers

Marcus looked up the path that he had taken on his run this morning on the map shown below.

AURORKA [14]

Answer: 6 miles.

Step-by-step explanation:

1. To solve this problem you only need to add the distances shown in the map attached, which shown the path Marcus had taken. Then, you have:

$Total_{distance}=1in+3inches+2inches+2.3inches\\Total_{distance}=8.3inches$

2. Now, you must convert the result obtained from inches to miles. According to the problem $4inches=3miles$ , therefore, you have:

$Total_{distance}=\frac{8,3inches*3miles}{4inches}\\Total_{distance}=6.22miles$

3. To the nearest mile:

$Total_{distance}=6miles$

6 0

4 years ago

Read 2 more answers

Jack likes to go fishing. While waiting for the fishes to bite, he formulates the following model for the process: fishes bite a

gogolik [260]

Answer:

a) 0.122

b) 3.35 × 10⁻⁴

c) 0.1071

Step-by-step explanation:

Here we have the Poisson process given by

$P(N(t) = k) = \frac{e^{-\lambda t} (\lambda t)^k }{k!}$

Where:

λ = Rate = 4 bites/hour and

t = Length of time that is

Mean = λt

Where k = 6 fishes bites and t = 2 hours we have

$P(N(2) = 6) = \frac{e^{-4 \times 2} (4 \times 2)^6 }{6!} = 0.122$

b) The probability that he fails to catch any fishes during the first two hours is the probability t that he catches 0 fish as follows

$P(N(2) = 0) = \frac{e^{-4 \times 2} (4 \times 2)^0 }{0!} = 3.35 \times 10^{-4}$

(c) Here we have

$P(N(2) = 6) = 0.122$

For 1 fish caught in two times we have first time no fish and second time he caches a fish gives

0.122×(1-0.122) = 0.1071

4 0

3 years ago