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3 years ago

Can U pls help with this number in the sequence

Mathematics

1 answer:

shtirl [24]3 years ago

7 0

Answer:

First one:

1², 2², 3² and so on (Square of 1,2,3,4,5,6,7.....)

Second one:

2, 13, 26, 41...

add 11, add 13, add 15 (add increasing odd numbers, beginning from 11)

Third one:

4, 14, 29, 49

+10, +15, +20 and so on (add the multiple of 5,beginning from 10)

Fourth one:

1, 8, 27, 64

1³, 2³, 3³, 4³ and so on(cube of 1,2,3,4,5,6,7.....)

Brainliest please

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2 years ago

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Arrange the functions in ascending order, starting with the function that eventually has the least value and ending with the fun

nataly862011 [7]

Answer:

Following are the responses to the given question:

Step-by-step explanation:

When x is squared, it's going to be bigger by the conclusion because the larger the faster. $x^2 + 4$ consistently exceeds $4 x^2$ , regardless of the price of X.

5x+3 is always 3 more so than 5x.

In particular, this can be classified according to its pathways $(5x$

$5x\\\\5x + 3\\\\7x\\\\8x + 3\\\\x^2\\\\x^2 + 4$

So can alternatively just say "eventually," which means because when x really is big, you need to find one another's values. Then just select x=100, then plug it into every single other. To confirm this order:

$5(100) = 500\\\\5(100) + 3 = 503\\\\7(100) = 700\\\\8(100) + 3 = 803\\\\(100)^2 = 10,000\\\\(100)^2 + 4 = 10,004\\\\$

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2 years ago

The odd- and even - number hotel rooms are on different sides of the hall. Room 231 is between witch two rooms

mario62 [17]

The first odd number before 231 which would be on the same side of the hall is 229. The first odd number after would be 233.
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3 years ago

Brad had a small gathering at a local steakhouse. The steakhouse offers three dinner platters which vary by size and price. They

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3 years ago

The probability that a male will be colorblind is .042. Find the probabilities that in a group of 53 men, the following are true

Dvinal [7]

Answer:

See below for answers and explanations

Step-by-step explanation:

Part A

$\displaystyle P(X=x)=\binom{n}{x}p^xq^{n-x}\\\\P(X=5)=\binom{53}{5}(0.042)^5(1-0.042)^{53-5}\\\\P(X=5)=\frac{53!}{(53-5)!*5!}(0.042)^5(0.958)^{48}\\\\P(X=5)\approx0.0478$

Part B

$P(X\leq5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\P(X\leq5)=\binom{53}{1}(0.042)^1(1-0.042)^{53-1}+\binom{53}{2}(0.042)^2(1-0.042)^{53-2}+\binom{53}{3}(0.042)^3(1-0.042)^{53-3}+\binom{53}{4}(0.042)^4(1-0.042)^{53-4}+\binom{53}{1}(0.042)^5(1-0.042)^{53-5}\\\\P(X\leq5)\approx0.9767$

Part C

$\displaystyle P(X\geq 1)=1-P(X=0)\\\\P(X\geq1)=1-(1-0.042)^{53}\\\\P(X\geq1)\approx1-0.1029\\\\P(X\geq1)\approx0.8971$

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2 years ago

Can U pls help with this number in the sequence​

Can U pls help with this number in the sequence