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3 years ago

A random number from 1 to 5 is selected

Mathematics

1 answer:

shutvik [7]3 years ago

7 0

Answer:

Relative frequency of selecting a 2 = 8/50 = 0.16

Relative frequency of selecting a 3 = 14/50 = 0.28

Step-by-step explanation:

When we have a given experiment with given outcomes (such that each time that we perform the experiment, one outcome happens) the relative frequency of a given outcome is the quotient between the number of times that that outcome happened, and the total number of times that the experiment was performed.

Here the experiment is selecting a random number between 1 and 5, and it is performed 50 times.

Out of these 50 times, the outcome "2" appears 8 times.

Then the relative frequency of selecting the number 2 is:

f(2) = 8/50 = 0.16

And of these 50 experiments, the outcome "3" appears 14 times.

Then the relative frequency of selecting the number 3 is:

f(3) = 14/50 = 0.28

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At an electronics store, a smart phone is on sale for 35% off the original price of $679. If you use the store credit card, you

kifflom [539]

Answer:

The final price of the smart phone if you use the store credit card is $339.5

Step-by-step explanation:

Given;

original price of the smart phone = $679

initial discount = 35%

additional discount if you use the store credit card = 15%

total discount of the smart phone if you use the store credit card

= 35% + 15% = 50%

The final price of the smart phone, if you use the store credit card is given as;

P = 0.5 X $679

P = $339.5

Therefore, the final price of the smart phone if you use the store credit card is $339.5

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3 years ago

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2 years ago

Find the product <br> .13 & 2.07 =<br> .73 & 1.5 =

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When the "product" is used in math it means to multiply so.

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3 years ago

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3 years ago

Read 2 more answers

1. Find four consecutive even integers such that the sum of the first and third is 6 less than the largest

KiRa [710]

1. Let the first integer be represented by: 2k

the second integer by: 2(k+1)

the third integer by: 2(k+2)

and the fourth by: 2(k+3)

Then, 2k + 2(k+2) = 2(k+3) - 6

2k + 2k + 4 = 2k + 6 - 6

4k + 4 = 2k

4 = -2k

-2 = k

Answer: -4, -2, 0, 2

******************************************************************

2. Let the first integer be represented by: 2k+1

the second integer by: 2k+3

the third integer by: 2k+5

and the fourth by: 2k+7

Then, 2k+7 = (2k+1) + (2k+3) + (2k+5) + 2

2k + 7 = 6k + 11

-4 = 4k

-1 = k

Answer: -1, 1, 3, 5

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3 years ago