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ki77a [65]
3 years ago
10

Need help on computer ‍

Mathematics
1 answer:
brilliants [131]3 years ago
7 0

Answer:

this is math not history but 32 29    32 29/40

                                                  40

Step-by-step explanation:

You might be interested in
16,42 greatest common factor
kozerog [31]
2

42 ÷ 2 = 21
16 ÷ 2 = 8

Let's try the other factors of 42...

42 ÷ 6 = 7
16 ÷ 6 = 2.66
No...

42 ÷ 7 = 6
16 ÷ 7 = 2 2/7
No...

42 ÷ 14 = 3
16 ÷ 14 = 1.1428
No...
7 0
3 years ago
Slips of paper numbered 1 to 15 are placed in a box. A slip of paper is drawn at random. What is the probability that the number
monitta

Given:

Slips of paper numbered 1 to 15 are placed in a box.

To find:

The probability that the number picked is either a multiple of 5 or an odd number.

Solution:

We have,

Total outcomes = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15

No. of total outcomes = 15

Multiple of 5 are 5, 10, 15.

Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15.

Number that are either a multiple of 5 or an odd number are 1, 3, 5, 7, 9, 10, 11, 13, 15.

No. of favorable outcomes = 9

We know that,

\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}

\text{Probability}=\dfrac{9}{15}

\text{Probability}=\dfrac{3}{5}

\text{Probability}=0.6

Therefore, the  probability that the number picked is either a multiple of 5 or an odd number is 0.6.

8 0
2 years ago
Find The Area of this<br> Triangle<br> 5 cm<br> 13 cm<br> 12 cm
nasty-shy [4]
The correct answer is 30cm
3 0
2 years ago
Yuhhhhhhhhhhhhhhhh I need help
baherus [9]

Answer:

JUST DO 3.142 X 10

ANSWER: 31.42 IN

Step-by-step explanation:

<em><u>Please mark as brainliest</u></em>

Have a great day, be safe and healthy

Thank u  

XD

4 0
2 years ago
A tobacco company claims that the amount of nicotine in its cigarettes is a randomvariable with mean 2.2 mg and standard deviati
Genrish500 [490]

Answer:

0.0000

Unusual

Step-by-step explanation:

Given that a tobacco company claims that the amount of nicotine in its cigarettes is a random variable with mean 2.2 mg and standard deviation .3 mg.

i.e. population parameters are

\mu =2.2 \\s = 0.3

The approximate probability that the sample meanwould have been as high or higher than 3.1

=P(X\geq 3.1)\\=P(Z\geq \frac{3.1-2.2}{\frac{0.3}{\sqrt{100} } } )\\=P(Z\geq 30)\\

=0.0000

8 0
3 years ago
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