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3 years ago

What is the binomial expansion of (x+1)^5

Mathematics

1 answer:

lana [24]3 years ago

4 0

Answer: "A"

somewhat of a trick question

use pascals triangle (or binomial expansion)

0-1

1-1,1

21,2,1

3- 1,3,3,1

4- 1,4,6,4,1

5- 1,5,10,10,5,1

NOTICE THAT "A" has the 1,5,10,10,5,1 in the 5th row

Step-by-step explanation:

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Can you please send me an example for this question <br> (15 ÷ 5 + 6 + 9) x 4 x 2 + 7

Mkey [24]

The answer is 88 hope that helped

3 0

3 years ago

Read 2 more answers

Farmers often sell fruits and vegetables at farmers’ markets during the summer. Each tomato stand at the Bentonville farmers’ ma

olga nikolaevna [1]

Answer:

the probability that all tomatoes are sold is 0.919 (91.9%)

Step-by-step explanation:

since the random variable X= number of tomatoes that are demanded, is normally distributed we can make the standard random variable Z such that:

Z=(X-μ)/σ = (83 - 125)/30 = -1.4

where μ= expected value of X= mean of X (since X is normally distributed) , σ=standard deviation of X

then all tomatoes are sold if the demand surpasses 83 tomatos , therefore

P(X>83) = P(Z>-1.4) = 1- P(Z≤-1.4)

from tables of standard normal distribution →P(Z≤-1.4)=0.081 , therefore

P(X>83) = 1- P(Z≤-1.4) = 1 - 0.081 = 0.919 (91.9%)

thus the probability that all tomatoes are sold is 0.919 (91.9%)

6 0

3 years ago

If u, v, and w are nonzero vectors in r 2 , is w a linear combination of u and v?

Tju [1.3M]

Not necessarily. $\mathbf u$ and $\mathbf v$ may be linearly dependent, so that their span forms a subspace of $\mathbb R^2$ that does not contain every vector in $\mathbb R^2$ .

For example, we could have $\mathbf u=(0,1)$ and $\mathbf v=(0,-1)$ . Any vector $\mathbf w$ of the form $(r,0)$ , where $r\neq0$ , is impossible to obtain as a linear combination of these $\mathbf u$ and $\mathbf v$ , since

$c_1\mathbf u+c_2\mathbf v=(0,c_1)+(0,-c_2)=(0,c_1-c_2)\neq(r,0)$

unless $r=0$ and $c_1=c_2$ .

8 0

3 years ago

It has long been reported that human body temperature follows a normal distribution with a mean of 98.6 degrees Fahrenheit. Ther

Kay [80]

Solution :

The normal body temperature of any human body is considered to be 98.6 $^\circ \ F$ . But there is a constant debate about the body temperature of a long held standard to the body temperature.