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3 years ago

Please answer i need help

Mathematics

1 answer:

r-ruslan [8.4K]3 years ago

8 0

Answer:

$\huge\boxed{\text{B) 1}}$

Step-by-step explanation:

In order to find which expression is equivalent to $(8^5 \times 3^{-7})^0$ , we need to note the order of operations.

PEMDAS tells us that when solving an equation, we'd go in this order.

<em>P</em><em>arentheses</em>

<em>E</em><em>xponents</em>

<em>M</em><em>ultiplication/</em><em>D</em><em>ivision</em>

<em>A</em><em>ddition/</em><em>S</em><em>ubtraction</em>

Using this, we'd start by evaluating what's inside the parentheses, $8^5 \times 3^{-7}$ . We'd then bring that value to the 0th power.

<em>Wait...</em> there's something special about bringing anything to the 0th power. Bringing anything to the 0th power makes the value 1.

It can be -5... it can be 1,000,000, any number possible, taken to the 0th power, will always be 1.

So no matter what the value inside the parentheses is, taking it to the 0th power will <u>always make it 1</u>.

Therefore, this expression is equivalent to 1.

Hope this helped!

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Answer:

a) P(X>np+3 $\sqrt{np(1-p)}$ =0.017

b) P(x>1)=0.190

c) P(Y>1)=0.651

Step-by-step explanation:

This a binomial experiment where success is denoted by parts that need rework.

X ∼ B(n, p); n = 20; p = 0.01

The expected value of X is: E(X) = np =20×0.01= 0.2

The variance is: Var(X) = np(1 − p) = 0.2 × 0.99 = 0.198,

The standard deviation SD(X)= $\sqrt{0.198}$ ≈ 0.445

a) P(X>np+3 $\sqrt{np(1-p)}$ =P(X>0.2+3×0.445)=P(X>1.535)=P(X≥2)

Probability function is given by:

$\frac{n!}{x!(n-x)!} *p^x*(1-p)^{(n-x)}$

P(X≥2)=1-P(X<2)=1-P(X=1)-P(X=0)= 1 - $\frac{20!}{1!(20-1)!} *(0.01)^{1}*(1-0.01)^{(20-1)}$ - $\frac{20!}{0!(20-0)!} *(0.01)^{0}*(1-0.01)^{(20-0)}$

P(X≥2)=1-0.165-0.818=0.017

b) p=0.04

P(x>1)=P(x≥2)= 1 - P(x=1) - P(x=0)= 1 - $\frac{20!}{0!(20-1)!} *(0.04)^{1}*(1-0.04)^{(20-1)}$ - $\frac{20!}{0!(20-0)!} *(0.04)^{0}*(1-0.04)^{(20-0)}$

P(x>1)= 1 - 0.368 - 0.442=0.190

c) In this case we consider p=0.19 (Probability that X exceeds 1)

In this experiment Y is the number of hours and n= 5 hours.

Then, we check the probability in each hour:

P(Y>1)=1- P(Y=0)

P(Y=0)= $\frac{5!}{0!(5-0)!} *(0.19)^{0}*(1-0.19)^{(5-0)}$ =0.349

P(Y>1)=1-0.349=0.651

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3 years ago

Please answer i need help​

Please answer i need help