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2 years ago

I have 3 pennies, 4 nickels, and 5 dimes. What ratio represents the following ratio: nickels to the total number of coins.

Mathematics

1 answer:

Karo-lina-s [1.5K]2 years ago

4 0

Answer:

1:3

Step-by-step explanation:

4:(3+4+5)

4:12

1:3

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3 years ago

39. The ratio of new houses to antique houses in a village is 4:5. If there are

jeyben [28]

Step-by-step explanation:

☄ $\underline{ \underline{ \text{Question }}}:$ The ratio of new houses to antique houses in a village is 4 : 5 . If there are 12 new houses , how many antique houses are there ?

☄ $\underline{ \underline{ \text{Solution}}} :$

✏ Let , the number of antique houses in a village be ' x ' .

Then , According to the question :

$\sf{ \frac{Number \: of \: new \: houses}{Number \: of \: antique \: houses} \: = \: \frac{4}{5}}$

⤑ $\sf{ \frac{12}{x} = \frac{4}{5}}$

Apply cross product property

⤑ $\sf{4 \times x = 12 \times 5}$

⤑ $\sf{4x = 60}$

⤑ $\sf{ \frac{4x}{4} = \frac{60}{4}}$

⤑ $\sf{x = 15}$

So, The number of antique houses in a village is 15.

$\red{ \boxed{ \boxed{ \tt{Our \: final \: answer : \boxed{ \bold{15}}}}}}$

Hope I helped ! ♡

Have a wonderful day / night ツ

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7 0

2 years ago

PLS HELP FOR 44 POINTS

Pie

Answer:

como es naturaleza HeS los de rosfentes lo 27

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3 years ago

A chef plans to mix 100% vinegar with Italian Dressing. The Italian dressing contains 8% vinegar. The chef wants to make 60 mill

GenaCL600 [577]

Answer:

Let v = ml of 100% vinegar

Then 150-v = ml of dressing

v + .05(150-v) = .24(150)

v + 7.5 - .05v = 36

.95v = 28.5

v = 28.5/.95

v = 30 ml of vinegar

dressing = 150-30 = 120 m

5 0

2 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

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

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

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

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

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

$P( X \le 4 ) = 0.0054$

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3 years ago