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2 years ago

Please answer this!!

Mathematics

1 answer:

Norma-Jean [14]2 years ago

8 0

Answer:

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Step-by-step explanation:

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The amount of time a certain brand of light bulb lasts is normally distributed with a

My name is Ann [436]

If anyone has seen the answers to this question then I need please post it

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2 years ago

Please help me with this question

Dmitrij [34]

Answer:

a) number 2 and 3 appeared a minimum number of times

b) number 5 appeared a maximum number of times

c) number 1, 4 and 6 appear an equal number of times (4 times). Also number 2 and 3 appear an equal number of times (1 time)

d) Even numbers turned up 9 times. The even numbers in a dice are 2 4 and 6. Add up how many times they came up: 1+4+4=9

f) Odd numbers turned up 11 times. The odd numbers in a dice are 1, 3 and 5. Add up how many times they came up: 4+1+6=11

7 0

2 years ago

PLEASE HELP I WILL GIVE BRAINLEST

Korvikt [17]

Answer:

Angle WZY=31degrees

Step-by-step explanation:

5x+11=8x-1

11+1=8x-5x

12=3x

X=12/3

X=4

Substitute x=4 in 5x+11

5(4)+11=31

6 0

3 years ago

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

3 years ago

A square classroom has a perimeter of 60 feet. How long is each side of the room?

nignag [31]

Answer:

Step-by-step explanation:

since it is a square all sides are equal.

60/4=15

Therefore the answer is 15

4 0

2 years ago

Read 2 more answers