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3 years ago

Can someone please help me with this and I need it today

Mathematics

1 answer:

viva [34]3 years ago

3 0

Answer:

Step-by-step explanation:

rule for line c: y=x+3

x y

0 3

1 4

2 5

3 6

I was able to determine the rule because line c has a slope (m) of 1 (it goes up one and sideways one each time). Its y int. is at 3, so that means 3 is b in the equation y= mx+b

Rule for line D: y=1/2x

x y

0 0

2 1

4 2

6 3

Line d is y=1/2x becaise it goes up 1 and then sidewys 2, so slope (m)= 1/2. There is no b becase it crosses the origin.

(-2,-1) (-4,-2) (-6, -3) just extend the line with a ruler.

hope this helps! :)

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_________% of 50 shirts is 35 shirts what is it

Rufina [12.5K]

Answer:

the answer is 70% hope that helps!!

6 0

2 years ago

Which fractions are equivalent to 10/12? Select each correct answer. 2/3 5/6 12/24 30/36 50/60 60/84

Yuri [45]

Answer:

[See Below]

Step-by-step explanation:

✦ Turn into decimals:

$\frac{10}{12}=0.83$

$\frac{2}{3} =0.67$

$\frac{5}{6} =0.83$

$\frac{12}{24} =0.5$

$\frac{30}{36} =0.83$

$\frac{50}{60} =0.83$

$\frac{60}{84} =0.71$

So $\frac{10}{12}$ ≈ $\frac{5}{6},\frac{30}{36},and,\frac{50}{60}$ .

~<em>Hope this helps Mate. If you need anything feel free to message me. </em>

4 0

3 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)

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