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2 years ago

Use the distributive property to write the following expression without parentheses. Then, simplify the result. -3(7x+3)+5=

Mathematics

1 answer:

Akimi4 [234]2 years ago

6 0

-3(7x+3)+5
=-21x-9+5
=-21x-4

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50 POINTS!!!!

Softa [21]

Answer:

Step-by-step explanation:

Given that X - the distribution of heights of male pilots is approximately normal, with a mean of 72.6 inches and a standard deviation of 2.7 inches.

Height of male pilot = 74.2 inches

We have to find the percentile

X = 74.2

Corresponding Z score = 74.2-72.6 = 1.6

P(X<174.2) = P(Z<1.6) = 0.5-0.4452=0.0548=5.48%

i.e. only 5% are below him in height.

Thus the malepilot is in 5th percentile.

3 0

1 year ago

Please help me with this expression problem!

kkurt [141]

Answer:

=k= -2.2/6.7

= k=(.328)

Step-by-step explanation:

2.2=k+6.7/-1

=2.2×-1=6.7+k

=k= -2.2/6.7

= k=(.328)

Hey mate hope it's help you.....

4 0

2 years ago

On a number line, point A has a coordinate of -6, and point B has a coordinate of 2. Which is the coordinate of point M, the mid

damaskus [11]

In my opinion
The answer is B

5 0

2 years ago

Which southern in their boxes shows has a greater value

suter [353]

Answer: C) They have equal volumes

Step-by-step explanation:

Volume of Rectangular Prism = length x width x height

Green: 4 x 2 x 2 =16

Purple: 8 x 2 x 1 = 16

8 0

2 years ago

Read 2 more answers

The rate of change of the number of mountain lions N(t) in a population is directly proportional to 725 - N(t), where t is the t

wolverine [178]

Answer:

a) The implied differential equation is $\frac{dN}{725 - N} = kdt$

b) The general equation is $N = 725 - C e^{-kt}$

c) The particular equation is $N = 725 - 325 e^{-0.49t}$

d) The population when t = 5, N(5) = 697 = 700( to the nearest 50)

Step-by-step explanation:

The rate of change of N(t) can be written as dN/dt

According to the question, $\frac{dN}{dt} \alpha (725 - N(t))$

$\frac{dN}{dt} = k (725 - N)\\\frac{dN}{725 - N} = kdt$

Integrating both sides of the equation

$\int {\frac{1 }{725 - N}} \, dN = \int {k} \, dt\\- ln (725 - N) = kt + C\\ ln (725 - N) = -(kt + C)\\725 - N = e^{-(kt + C)} \\725 - N = e^{-kt} * e^{-C} \\725 - N = C e^{-kt}\\N = 725 - C e^{-kt}$

When t = 0, N = 400

$400 = 725 - C e^{-k*0}\\400 = 725 - C\\C = 725 - 400\\C = 325$

When t = 3, N = 650

$650 = 725 - (325 * e^{-3k})\\325 * e^{-3k} = 75\\e^{-3k} = 75/325\\e^{-3k} = 0.23\\-3k = ln 0.23\\-3k = -1.47\\k = 1.47/3\\k = 0.49$

The equation for the population becomes:

$N = 725 - 325 e^{-0.49t}$

At t = 5, the population becomes:

$N = 725 - 325 e^{-0.49*5}\\N = 725 - 325 e^{-2.45}\\N = 696.95\\N(5) = 697$

N(5) = 700 ( to the nearest 50)

6 0

3 years ago