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3 years ago

Plz help , due in 5 mins

Mathematics

1 answer:

Grace [21]3 years ago

3 0

Answer:

option a) 30 : 19

Step-by-step explanation:

$Length = 7 \frac{1}{2} = \frac{15}{2}\\\\Width = 4\frac{3}{4} = \frac{19}{4}$

Ratio of length to width:

$\frac{15}{2} : \frac{19}{4}$

$=\frac{\frac{15}{2}}{\frac{19}{4}}\\\\=\frac{15}{2 } \times \frac{4}{19}\\\\=15\times \frac{2}{19}\\\\=\frac{30}{19}\\\\30 : 19$

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The average weekly income of teachers in one school district is $750 with a standard deviation of $85. Find the percentage of te

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Point estimate (average) = $750
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4 years ago

How many ten-digit numbers have at least two equal digits?

hammer [34]

Answer:

8,996,734,079 numbers

Step-by-step explanation:

First let's find the number of ten-digit numbers.

The maximum ten-digit number is 9,999,999,999 and the minimum is 1,000,000,000. So the number of ten-digit numbers is:

$N(total) = 9999999999 - 1000000000 = 8,999,999,999$

Now, to find the number of ten-digit numbers with at least two equal digits, we can find the number of ten-digit numbers with all digits different, and then subtract this amount from the total ten-digit numbers.

To find the number of ten-digit numbers with all digits different, we can use the following logic:

The first digit can have 9 different values (0 not included), the second can also have 9 (one digit used), then the third can have 8 (two digits used), the fourth can have 7, and so on. So we have that:

$N(different) = 9*9! = 3,265,920$

Then the number of ten-digit numbers with at least two equal digits is:

$N(two\ equal) = N(total)-N(different) = 8999999999 - 3265920$

$N(two\ equal) = 8,996,734,079$

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3 years ago

Plz help , due in 5 mins ​

Plz help , due in 5 mins