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3 years ago

I need help please please

Mathematics

1 answer:

torisob [31]3 years ago

7 0

The correct answer to this question is false.

The number 0.8 can in fact be written as 8/10 (because when you divide 8/10 you get 0.8 so they are equivalent values); however, rational numbers are those that can be written as a fraction (the division of two numbers). Irrational numbers are numbers that cannot be expressed in fractional form. This means that 8/10 is not irrational, it is rational, because you can express it in fractional form. This means your answer is false.

Hope this helps!

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Which number is not equal to one of the following expressions

Papessa [141]

Answer:150,000

Step-by-step explanation:

1500×10=15000

150,000÷100=1500

1500÷10=150

So 150,000 is not an answer in one the expressions

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2 years ago

Ken paid for $12 for two magazines. The cost of each magazine was a multiple of 3. What is Mare the possible prices of the magaz

Artist 52 [7]

Answer:

So the possibilities are

$12 and $ 0 , $9 and $ 3 , $6 and $6

Step-by-step explanation:

Given:

Total Magazines = 2

Price = $ 12

To Find:

Possible Prices of magazines = ?

Solution:

As it is given that Prices of magazines is multiples of 3

Multiples of 3 up till 12 are

0, 3 , 6 , 9 , 12

Now

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Now The possible prices of the magazines are

First Possibility:

If price of first magazine is $12 then second would be $0 (free)

First possible price $12 and $0

Second Possibility:

If price of first magazine is $9 then second would be $3

Second possible price $3 and $9

Third Possibility:

If price of first magazine is $6 then second would be $6

Third possible price $6 and $6

So the possibilities are

$12 and $ 0 , $9 and $ 3 , $6 and $6

8 0

3 years ago

The probability that a male will be colorblind is .042. Find the probabilities that in a group of 53 men, the following are true

Dvinal [7]

Answer:

See below for answers and explanations

Step-by-step explanation:

Part A

$\displaystyle P(X=x)=\binom{n}{x}p^xq^{n-x}\\\\P(X=5)=\binom{53}{5}(0.042)^5(1-0.042)^{53-5}\\\\P(X=5)=\frac{53!}{(53-5)!*5!}(0.042)^5(0.958)^{48}\\\\P(X=5)\approx0.0478$

Part B

$P(X\leq5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\P(X\leq5)=\binom{53}{1}(0.042)^1(1-0.042)^{53-1}+\binom{53}{2}(0.042)^2(1-0.042)^{53-2}+\binom{53}{3}(0.042)^3(1-0.042)^{53-3}+\binom{53}{4}(0.042)^4(1-0.042)^{53-4}+\binom{53}{1}(0.042)^5(1-0.042)^{53-5}\\\\P(X\leq5)\approx0.9767$