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3 years ago

What is the estimate of 615

Mathematics

2 answers:

Yakvenalex [24]3 years ago

8 0

Estimate to nearest hundreds is 600

Estimate to nearest tens is 610 or 520

Artemon [7]3 years ago

8 0

600 or 620. Hope this helps!

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What is the fourth term in the binomial expansion (a+b)^6)

Dafna11 [192]

Answer:

$20a^3b^3$

Step-by-step explanation:

<u>Binomial Series</u>

$(a+b)^n=a^n+\dfrac{n!}{1!(n-1)!}a^{n-1}b+\dfrac{n!}{2!(n-2)!}a^{n-2}b^2+...+\dfrac{n!}{r!(n-r)!}a^{n-r}b^r+...+b^n$

<u>Factorial</u> is denoted by an exclamation mark "!" placed after the number. It means to multiply all whole numbers from the given number down to 1.

Example: 4! = 4 × 3 × 2 × 1

Therefore, the fourth term in the binomial expansion (a + b)⁶ is:

$\implies \dfrac{n!}{3!(n-3)!}a^{n-3}b^3$

$\implies \dfrac{6!}{3!(6-3)!}a^{6-3}b^3$

$\implies \dfrac{6!}{3!3!}a^{3}b^3$

$\implies \left(\dfrac{6 \times 5 \times 4 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}{3 \times 2 \times 1 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}\right)a^{3}b^3$

$\implies \left(\dfrac{120}{6}\right)a^{3}b^3$

$\implies 20a^3b^3$

7 0

2 years ago

If 75% is 10 days, how many days is the remaining 25%?

inna [77]

Answer:

10/3 days or 3 1/3 days

Step-by-step explanation:

Let d = number of days total

75% of the days is 10

.75 d = 10

d = 10/.75

d = 40/3 days

We want to know how man is 25% of d

d * 25%

(40/3) *.25

10/3 days

6 0

2 years ago

John is playing a game with a standard deck of playing cards. He wants to draw a jack on the first try. Which of the following s

Tatiana [17]

Answer:

Step-by-step explanation:

It is a 1/13 chance but even if you shuffle and replace it, its still a 1/13 chance.

4 0

3 years ago

Read 2 more answers

HELPPPPPP!!

AlekseyPX

Answer:

In 6 different ways can the three students form a set of class officers.

Step-by-step explanation:

There are 3 people Leila, Larry, and Cindy and 3 positions president, vice-president, and secretary.

We need to find In how many different ways can the three students form a set of class officers.

This problem can be solved using Permutation.

nPr = n!/(n-r)! is the formula.

Here n = 3 and r =3

So, 3P3 = 3!/(3-3)!

3P3 = 3!/1

3P3 = 3*2*1/1

3P3 = 6

So, in 6 different ways can the three students form a set of class officers.

7 0

3 years ago

10÷7(19+28)×20÷6=???

Snezhnost [94]

Answer:

109 2/3

Step-by-step explanation:

10 / 7(47) x 20 / 6 =

10 / 329 x 20 / 6 =

32.9 x 20 / 6 =

658 / 6 =

109 2/3

Hope that helps!

6 0

3 years ago