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mariarad [96]
3 years ago
12

Who has longer thumbs explain how you know

Mathematics
1 answer:
gizmo_the_mogwai [7]3 years ago
6 0
<u><em>Boiz Body Are larger than girls,And a boi is taller then a girl</em></u>
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Roberto has eight coins that are all dimes or nickels. They are worth $0.50. How many dimes and how many nickels does Roberto ha
natita [175]

Answer: Roberto has 2 DIMES and 6 NICKELS.

Step-by-step explanation:

<em><u>Equation 1:</u></em> 2 * 10 = 20 cents (Dimes)

<em><u>Equation 2:</u></em> 6 * 5 = 30 cents (Nickels)

<em> </em><u><em>Solution:</em></u><em> </em>30 + 20 = 50 cents

7 0
3 years ago
Please help and if you dont know please don't answer
Allisa [31]

Answer:

81cm

Step-by-step explanation:

15cm + 20cm + 21cm + 25cm = 81cm

5 0
2 years ago
Read 2 more answers
A test consists of 10 multiple choice questions, each with five possible answers, one of which is correct. To pass the test a st
Softa [21]

Answer:

0.006369

Step-by-step explanation:

Given that a test consists of 10 multiple choice questions, each with five possible answers, one of which is correct.

By mere guessing p = probability for a right answer = 1/5 =0.20

There are two outcomes and each question is independent of the other.

X no of questions right is Bin (10,0.20)

the probability that the student will pass the test

= prob of getting more than 60%

=P(X\geq 6)

=0.006369

8 0
3 years ago
Use mathematical induction to prove
Alex17521 [72]

Prove\ that\ the\ assumption \is \true for\ n=1\\1^3=\frac{1^2(1+1)^2}{4}\\ 1=\frac{4}{4}=1\\

Formula works when n=1

Assume the formula also works, when n=k.

Prove that the formula works, when n=k+1

1^3+2^3+3^3...+k^3+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4} \\\frac{k^2(k+1)^2}{4}+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4} \\\frac{k^2(k^2+2k+1)}{4}+(k+1)^3=\frac{(k^2+2k+1)(k^2+4k+4)}{4} \\\frac{k^4+2k^3+k^2}{4}+k^3+3k^2+3k+1=\frac{k^4+4k^3+4k^2+2k^3+8k^2+8k+k^2+4k+4}{4}\\\\\frac{k^4+2k^3+k^2}{4}+k^3+3k^2+3k+1=\frac{k^4+6k^3+13k^2+12k+4}{4}\\\frac{k^4+2k^3+k^2}{4}+\frac{4k^3+12k^2+12k+4}{4}=\frac{k^4+6k^3+13k^2+12k+4}{4}\\\frac{k^4+6k^3+13k^2+12k+4}{4}=\frac{k^4+6k^3+13k^2+12k+4}{4}\\

Since the formula has been proven with n=1 and n=k+1, it is true. \square

7 0
2 years ago
Which equation could be used to solve the question below?
ivann1987 [24]

It would be useful if we knew what the equation is

8 0
3 years ago
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