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2 years ago

I’m doing a test can anyone tell me what to click ?

Mathematics

1 answer:

vova2212 [387]2 years ago

4 0

Answer:

the bottom right

Step-by-step explanation:

6/8= 75%

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What would the bottles purchased be and what would the bottles recycled be?

N76 [4]

Part A

Your problem statement already shows you the conversion.

Bottles purchased: 5×10¹⁰

Bottles recycled: 8.3×10⁹

Part B

The ratio of bottles purchased to bottles recycled is

... (5×10¹⁰)/(8.3×10⁹) = 0.6×10¹ = 6.0

The number of bottles purchased is 6 times the number of bottles recycle.

5 0

2 years ago

Resolva esta questão usando a lei do anulamento 2×(×-5) =0

Kay [80]

<span>Hola, hablas español?</span>

6 0

2 years ago

The sum of two numbers is 17, and their difference is 29. What are the two numbers?

serious [3.7K]

X + y = 17
x - y = 29

x = 17 - y
x - y = 29

x = 17 - y
17 - y - y = 29

x = 17 - y
17 - 2y = 29

x = 17 - y
- 2y = 29 - 17

x = 17 - y
-2y = 12

x = 17 - y
y = -6

x = 17 - (-6)
y = -6

x = 17 + 6
y = -6

x = 23
y = -6

6 0

2 years ago

Read 2 more answers

I'm lazy and dont feel like doing math myself. So pls help me!

s2008m [1.1K]

Answer:

Step-by-step explanation:

Well you just plug in 10 as n.. and you get the equation:

$\frac{10(10+1)}{2} \\\frac{10(11)}{2}\\\\\frac{110}{2}\\\\55$

8 0

1 year ago

A computer programmer has an 18% chance of finding a bug in any given program. What is the probability that she does not find a

LiRa [457]

Answer: $(\frac{41}{50})^{10}$

Step-by-step explanation:

Since, According to the question,

The chance of finding a bug in a program = 18%

Thus, the probability of finding the bug in first attempt = $\frac{18}{100}=\frac{9}{50}$

⇒ The probability of not finding any bug in first attempt = $1 - \frac{9}{50} =\frac{41}{50}$

Similarly, in second attempt , third attempt, fourth attempt_ _ _ _ _ tenth attempt, the probability of not finding any bug is also equal to $\frac{41}{50}$

Thus, the probability that she does not find a bug within the first 10 programs she examines

= $\frac{41}{50}\times \frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}$

= $(\frac{41}{50})^{10}$

5 0

2 years ago