What is the 20 digit of 89

Because of the commutative property of multiplication, it is true that

rewona [7]

Answer:

It is Commutative

Step-by-step explanation:

An operation ∆ is said to be Commutative if a∆b=b∆a ∀ a,b ∈ ℝ.

Given the operation ∆ defined by:

a∆b=a X b

$a=\frac{3}{4}, b=4$

a∆b= $\frac{3}{4}X4$

$=\frac{3X4}{4}$ =3

Similarly, for the right hand side.

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

Therefore:

b∆a= $4X \frac{3}{4}$

$=\frac{4X3}{4}$ =3

These are the two ways of solving this problem and we have in fact shown that the operation is commutative as:

a∆b=b∆a=3

3 0

3 years ago

A hot air balloon rises an initial 525

ludmilkaskok [199]

Answer:

380ft

Step-by-step explanation:

Assuming the balloon rises 525ft from sea level (0ft), all we need to do is subtract both numbers to find the new height.

525 - 145 = 380ft

There could be many different scenarios though because if the question provides the height of land off sea level, you would have to take that into account as well.

Best of Luck!

5 0

2 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .