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3 years ago

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Mathematics

1 answer:

lara [203]3 years ago

8 0

$\frac{7x + 42}{x^{2} + 13x +42} = \frac{7(x+6)}{(x + 6)(x + 7)} = \frac{7}{x + 7}$

$Answer C.) \frac{7}{x+7}$

$\textbf{Spymore}$

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The distance of a train from a station, varies directly with time, t. If d = 300 miles when t = 4 hours, find d when t = 6.

mart [117]

Answer:

450 miles

Step-by-step explanation:

there is a 2 hr difference in betwen them. when you take 450 there is a 150 mile difference from 300 so when you divide 150 by 2 you get 75 so you do about 75 miles an hr so you can go 450 miles in 6 hr's.

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7B5%7D%20%20-%207%20%3D%204" id="TexFormula1" title=" \frac{x}{5} - 7 = 4"

Anestetic [448]

Hello! And thank you for your question!

First add 7 to both sides:

x/5 = 4 + 7

Then simplify 4 + 7:

x/5 = 11

After that, multiply both sides by 5

11 x 5 = x

Finally, Simplify 11 * 5:

x = 55

Final Answer:

x = 55

3 0

4 years ago

What is the slope of the line through (3, 3) and (0, 12)? Pls help due soon

Lapatulllka [165]

Answer:

y=-3x+12 equation of line

however slope is -3

Step-by-step explanation:

y=mx+c

(12-3)/(0-3)=9/-3=-3

y=-3x+c

3=-3(3)+c

3=-9+c

12=c

y=-3x+12

3 0

3 years ago

A package of 3 notebooks cost $5. Complete the ratio table and graph the pairs of values. How much will 18 notebooks cost?

victus00 [196]

Answer:

$30

Step-by-step explanation:

18÷3=6 (because the notebooks come in packages of 3)

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7 0

3 years ago

The annual rainfall (in inches) in a certain region is normally distributed with µ = 40 and σ = 4. What is the probability that

Allushta [10]

Answer:

P( That it will take over 10 years or more of a year with a rainfall above 50inches) = (0.9938)^10

Step-by-step explanation:

Since the annual rainfall is normally distributed,

Given: that

Mean (µ )= 40

and σ = 4.

Let X be normal random variables of the annual rainfall.

P(that there will be over 10 years or more before a year with a rainfall above 50 inches)

P(>50) = 1-P[X ≤50]

1 - P[X- μ/σ ≤ 50 - 40/4]

=1 - P [Z≤ 5/2]

=1 -Φ(5/2)

=1 - 0.9939

= 0.0062

P( the non occurrence of rainfall above 50 inches)

= 1-0.0062

=0.9938

ASSUMPTION:

P( That it will take over 10 years or more of a year with a rainfall above 50inches) = $(0.9938)^10$

4 0

3 years ago