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3 years ago

Unit 1 Lesson 4 Quiz

Mathematics

1 answer:

Arisa [49]3 years ago

5 0

Answer:

Sorry man but I can't see the question : (

Step-by-step explanation:

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Question 1 (1 point)

Firdavs [7]

Answer:

B. false

Step-by-step explanation:

Square units

I hope that is useful for you :)

6 0

3 years ago

There are 4 blue, 2 red, 1 brown, 1 black, 1 white, and 3 green crayons in a box. What is the probability of picking a red crayo

loris [4]

Answer:

1/6

Step-by-step explanation:

Simply just Add the different amounts of different crayons as your denominator, and find the number of red, as your numerator. This gets you 2/12. 2/12 = 1/6

Can I be brainliest? TYSM

8 0

2 years ago

Read 2 more answers

Twenty percent of drivers driving between 10 pm and 3 am are drunken drivers. In a random sample of 12 drivers driving between 1

Lesechka [4]

Answer:

(a) 0.28347

(b) 0.36909

(d) 0.9806

Step-by-step explanation:

Given information:

n=12

p = 20% = 0.2

q = 1-p = 1-0.2 = 0.8

Binomial formula:

$P(x=r)=^nC_rp^rq^{n-r}$

(a) Exactly two will be drunken drivers.

$P(x=2)=^{12}C_{2}(0.2)^{2}(0.8)^{12-2}$

$P(x=2)=66(0.2)^{2}(0.8)^{10}$

$P(x=2)=\approx 0.28347$

Therefore, the probability that exactly two will be drunken drivers is 0.28347.

(b)Three or four will be drunken drivers.

$P(x=3\text{ or }x=4)=P(x=3)\cup P(x=4)$

$P(x=3\text{ or }x=4)=P(x=3)+P(x=4)$

Using binomial we get

$P(x=3\text{ or }x=4)=^{12}C_{3}(0.2)^{3}(0.8)^{12-3}+^{12}C_{4}(0.2)^{4}(0.8)^{12-4}$

$P(x=3\text{ or }x=4)=0.236223+0.132876$

$P(x=3\text{ or }x=4)\approx 0.369099$

Therefore, the probability that three or four will be drunken drivers is 0.3691.

(c)

At least 7 will be drunken drivers.

$P(x\geq 7)=1-P(x$

$P(x\leq 7)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)]$

$P(x\leq 7)=1-[0.06872+0.20616+0.28347+0.23622+0.13288+0.05315+0.0155]$

$P(x\leq 7)=1-[0.9961]$

$P(x\leq 7)=0.0039$

Therefore, the probability of at least 7 will be drunken drivers is 0.0039.

(d) At most 5 will be drunken drivers.

$P(x\leq 5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)$

$P(x\leq 5)=0.06872+0.20616+0.28347+0.23622+0.13288+0.05315$

$P(x\leq 5)=0.9806$

Therefore, the probability of at most 5 will be drunken drivers is 0.9806.

5 0

3 years ago

2/7m-1/7=3/14<br><br> Please solve showing work

lawyer [7]

(2/7)m - (1/7) = 3/14
2m/7 =(3/14) + (1/7)
2m/7 = (3/14) + 2(1/7)
here we are multiplying 2 with 1/7 to make the denominator same for addition.
2m/7 = (3/14) +(2/14)
2m/7 = (3 + 2)/14
2m/7 = 5/14
2m = (5 *7)/14
2m = 35/14
2m = 5/2
m = 5/4
m = 1.25
So the value of "m" is 1.25

7 0

3 years ago

Read 2 more answers

Write a division problem whose quotient has first digit in the hundreds place

AysviL [449]

OK.

Divide 1,000 by 10.

7 0

3 years ago