What number needs to be added to 5 and 3 so that the ratio of the first number to the second

Mathematics

1 answer:

Mashcka [7]2 years ago

5 0

Answer:

Step-by-step explanation:

Choose a variable to represent the number to be added to 5 and 3.

$\frac{5+x}{3+x}=\frac{4}{3}$

"Cross-multiply."

$3(5+x)=4(3+x)\\\\15+3x=12+4x\\\\15=12+x\\\\3=x$

Check: The ratio of 5+3 to 3 + 3 is 8:6, equal to the ratio 4:3.

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If _______ , what is f^-1(x)?<br> f–1(x) = 9x + 18<br> ________<br> f–1(x) = 9x + 2<br> ________

Dennis_Churaev [7]

If f(x)=1/9x-2
then f^-1(x) will be
lets assume f(x)=y
y=1/9x-2
y+2=1/9x
9(y+2)=x
f^-1(x)=9(y+2)

rest u can do with the same method

5 0

4 years ago

Read 2 more answers

A truck driver can travel 560 miles on 28 gallons of gas. How far can he travel on 35 gallons of gas?

Shtirlitz [24]

560 / 28 = 20....so the driver travels 20 miles per gallon

on 35 gallons......35 * 20 = 700 miles <==

or u can also do it this way...

560/28 = x / 35....560 miles to 28 gal = x miles to 35 gal
cross multiply because this is a proportion
(28)(x) = (560)(35)
28x = 19600
x = 19600/28
x = 700 miles <==

so u can either do it by finding the unit rate or by making a proportion. either way will result in the same answer.

8 0

4 years ago

At an airport, 76% of recent flights have arrived on time. A sample of 11 flights is studied. Find the probability that no more

I am Lyosha [343]

Answer:

The probability is $P( X \le 4 ) = 0.0054$

Step-by-step explanation:

From the question we are told that

The percentage that are on time is p = 0.76

The sample size is n = 11

Generally the percentage that are not on time is

$q = 1- p$

$q = 1- 0.76$

$q = 0.24$

The probability that no more than 4 of them were on time is mathematically represented as

$P( X \le 4 ) = P(1 ) + P(2) + P(3) + P(4)$

=> $P( X \le 4 ) = \left n } \atop {}} \right.C_1 p^{1} q^{n- 1} + \left n } \atop {}} \right.C_2p^{2} q^{n- 2} + \left n } \atop {}} \right.C_3 p^{3} q^{n- 3} + \left n } \atop {}} \right.C_4 p^{4} q^{n- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{11- 1} + \left 11 } \atop {}} \right.C_2p^{2} q^{11- 2} + \left 11 } \atop {}} \right.C_3 p^{3} q^{11- 3} + \left 11 } \atop {}} \right.C_4 p^{4} q^{11- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{10} + \left 11 } \atop {}} \right.C_2p^{2} q^{9} + \left 11 } \atop {}} \right.C_3 p^{3} q^{8} + \left 11 } \atop {}} \right.C_4 p^{4} q^{7}$

$= \frac{11! }{ 10! 1!} (0.76)^{1} (0.24)^{10} + \frac{11!}{9! 2!} (0.76)^2 (0.24)^{9} + \frac{11!}{8! 3!} (0.76)^{3} (0.24)^{8} + \frac{11!}{7!4!} (0.76)^{4} (0.24)^{7}$