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3 years ago

Will give brainliest!!

Mathematics

2 answers:

rodikova [14]3 years ago

4 0

The answer is option C) 7.688 * 10^8.

It's simple. Just add 5 and 3. Then you will get 8. And then multiply 3.844 * 2 . You will get 7.688. After that write this down : 7.688 * 10^8.

Hope this answer will help you a lot. If it does then mark my answer as the brainliest please:)!

MakcuM [25]3 years ago

4 0

Answer: Option C.

Step-by-step explanation:

Average distance from Earth to the Moon (d1) = 3.844*10^5 km.

Average distance from Jupiter to the Sun (d2) = 2*10^3 times the average distance from Earth to the Moon (d1).

Which is same as

d2 = 2*10^3 × d1

d2 = 2*10^3 × 3.844*10^5

d2 = 2*3.844*10^3*10^5

d2 = 7.688*10^(3+5)

d2 = 7.688*10^8

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A developer bought 78 trees at a total cost of $4050. The shade trees cost $45 each and the evergreens cost $60 each. How many o

ANTONII [103]

Answer:

Step-by-step explanation:

for 5 feet is $60

2 feet is $150.

sorry if this doesn't help!

6 0

2 years ago

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

Help help help help

deff fn [24]

Answer:

b = 12

Step-by-step explanation:

b/3 + 4 = 8

subtract 4 on both sides

b/3 = 4

multiply 3 with 4

b = 12

4 0

2 years ago

Área de Cuadriláteros <br> ¿Me pueden ayudar con esta tarea?

IRISSAK [1]

Answer:

I can't understand this language sorry

7 0

2 years ago

Math: Slope help please

pishuonlain [190]

Slope = (2+5)/(-3-1) = 7/-4 = -7/4

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3 0

3 years ago

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