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2 years ago

A single die is rolled twice. Find the probability of rolling an

Mathematics

1 answer:

Nady [450]2 years ago

6 0

Answer: 1/6

Step-by-step explanation:

A die has 6 numbers which are 1, 2, 3, 4, 5 and 6.

Odd numbers in a die = 1, 3 and 6

Numbers greater than 4 = 5 and 6

Probability of rolling an odd number = 3/6 = 1/2

Probability of rolling a number greater than 4 = 2/6 = 1/3

We then multiply both values gotten. This will be:

= 1/2 × 1/3

= 1/6

Therefore, the probability of rolling an odd number the first time and a number greater than 4 the second time is 1/6.

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Ng 20-21

Lisa [10]

you would spend $30

Step-by-step explanation:

if 20 calculator cost 120 then you would divide 20 by 120 which would give you 6 dollars per calculator then you would multiply that by 5 which would give you a $30 for five calculators.

4 0

2 years ago

Read 2 more answers

your gym membership cost $33 per month after an initial membership fee. you paid a total of $228 after 6 months. writer an equat

grin007 [14]

Answer:

what i did was multiply 33*6 and got 198 and then 228-198 to find out that the initial fee was 30 dollars. then multiply 33*9+30 to get a 9-month total cost of $327

Step-by-step explanation:

cost = $33 per month + fee

1. c(m) = 33m + f

33(6) + f = 228

198 + f = 228

f = $30 membership fee

c(m) = 33m + 30

2. at 9 months: 33(9) + 30 = $327

hopefully this helps :)

have a nice day !!

7 0

2 years ago

Ill give brainliest <br> could someone please help me

polet [3.4K]

Answer: 23.2

Step-by-step explanation:

8 0

2 years ago

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$