Helpmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

Mathematics

1 answer:

alexandr402 [8]2 years ago

6 0

Answer:

B. 1.2×10^-3 mm

Step-by-step explanation:

What constitutes an "appropriate" unit depends on the use of the number. Since we're not told of any particular use, we might choose the smallest of the available units: mm.

1.2×10^-3 mm

_____

<em>Additional comment</em>

The most appropriate unit would be the one that eliminates the power-of-ten multiplier:

= 1.2 µm

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Use the graph of f below. Assume the entire function is graphed below.<br> Find the domain

enot [183]

ANSWER

The domain of the function is

$- 3 \leqslant x \leqslant 6$

[-3,6]

EXPLANATION

The domain of a function refers to the values of x for which the function is defined.

From the graph, we can see that the function begins at (-3,6) and ends at (6,-4).

Therefore the function is defined for x=-3 to x=6.

Hence the domain of the function is

$- 3 \leqslant x \leqslant 6$

[-3,6]

5 0

2 years ago

An automated egg carton loader has a 1% probability of cracking an egg, and a customer will complain if more than one egg per do

vaieri [72.5K]

Answer:

a) Binomial distribution B(n=12,p=0.01)

b) P=0.007

c) P=0.999924

d) P=0.366

Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

$P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107$

Then,

$P(k>1)=1-(P(k=0)+P(k=1))=1-(0.886+0.107)=1-0.993=0.007$

c) In this case, the distribution is B(1200,0.01)

$P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924$