## What should I feed my 16hh Standardbred?

I have a 16hh standardbred that is 9 years old I got him about 6 months ago and its coming close to winter, I want to know what I should feed him? He is kept on grass and is ridden 4-7 times a week mainly walking and trotting but sometimes the odd canter… I will take any suggestions (Im thinking about something like chaff and ???)

Feeding isn’t dependent on his height, but his weight. Here is a site that gives you an approximate estimate on how much you should be feeding him.

Http://www.baileyshorsefeeds.co.uk/feedingexplained/calculator.htm

## How can I find the possible combinations on a 9-digit number that contain all the odd numbers(0-9)?

I have found out that the possible combinations on a 9-digit number with 10 variables (0-9) are 10^9 but how can I find how many only the combinations that include all odd numbers (e.g. 773091258) are?

I will assume that 0 is allowed as a first digit (otherwise there would be less than 10^9 possible 9-digit numbers, contrary to what is stated in the question).

Now we count how many 9-digit numbers include all five odd digits 1, 3, 5, 7, 9. First, we find the number of 9-digit numbers that exclude at least one of the odd digits, by using the inclusion-exclusion principle, then subtract the result from the 10^9 possible 9-digit numbers.

Note the following:

9^9 numbers exclude a given odd digit;
8^9 numbers exclude two given odd digits;
7^9 numbers exclude three given odd digits;
6^9 numbers exclude four given odd digits; and
5^9 numbers exclude all five odd digits.

By the inclusion-exclusion principle, the number of 9-digit numbers that exclude at least one odd digit is

5C1(9^9) – 5C2(8^9) + 5C3(7^9) – 5C4(6^9) + 5C5(5^9).

So the number of 9-digit numbers that include all five odd digits is

10^9 – 5C1(9^9) + 5C2(8^9) – 5C3(7^9) + 5C4(6^9) – 5C5(5^9)
= 10^9 – 5(9^9) + 10(8^9) – 10(7^9) + 5(6^9) – 5^9
= 49,974,120, using a calculator.

This means that about 5% of the 10^9 nine-digit numbers include all the odd digits.

Addendum: I now see that another answerer is assuming that 0 is not allowed as a first digit. In that case, the number of nine-digit numbers that include all the odd digits is modified to

9*10^8 – 5C1(8*9^8) + 5C2(7*8^8) – 5C3(6*7^8) + 5C4(5*6^8) – 5C5(4*5^8)
= 9*10^8 – 5(8*9^8) + 10(7*8^8) – 10(6*7^8) + 5(5*6^8) – 4*5^8
= 47,076,120, using a calculator.

This means that of the 9*10^8 nine-digit numbers with nonzero first digit, about 5.23% include all of the odd digits.

## If there is a 50% chance that I hit something, what are the odds I will hit it seven times in a row ?

That is if I only shoot seven times…

how do the odds increase of it happening if I shoot at it seventy times ?

If you only get 7 shots, the odds are:
.5 ^ 7 = 0.0078125 = 0.78%

If you want to hit something 70 times in a row:
.5 ^ 70 = 8.47033E-22 = 8.47033E-20 % (isn’t going to happen)

Now, the INTERESTING problem (which I’m pretty sure you did ask) is: If you get 70 shots, what are the odds you can hit your target 7 times a row at any point during your attempt. They should improve. This means you get 64 attempts at the 7 times in a row problem (Once you get within the last 7 tries, if you haven’t succeeded by then, you aren’t going to). You really need to calculate the odds of NOT hitting it 7 times in a row and then subtract that from 1:

odds of not hitting it 7 times in a row: 1-0.007813
odds of doing that 64 times: (1-0.007813)^64
odds of NOT doing that 64 times:
(1 – (1-0.007813)^64) = 0.394659009 = 39.5%

——————————————
And just for fun:

If you simply want to hit it at least 7 times and you get 70 shots, then you first need to find out how many outcomes there are (2 ^ 70) and how many of those outcomes have 7 or more hits. This is a combinatorics problem and is easier if you find how many outcomes do not have at least 7 hits.

C(0) = 1 arrangement where there are NO hits
C(1) = 70 arrangements where there is 1 hit
C(2) = 69*69 arrangements where there are 2 hits
C(3) = 68^3 for 3 hits
C(4) = 67^4 for 4 hits
C(5) = 66^5 for 5 hits
C(6) = 65^6 for 6 hits

Sum(C(0:6)) = 76,691,693,586 possible configurations in which fewer than 7 hits occurred in 70 attempts.

Soooo… 76,691,693,586 / 2^70 is approximately 6.5E-11. So it’s highly improbable. Most calculators will give 1 minus that number as 1. (The mantissa was too big).

## What are the odds of getting hired as a radiation therapist?

good or bad? how much do they get paid?

I have some friends who had a horrible time getting a job. Others got hired pretty fast. They get paid VERY WELL considering you’ll have either a BS or a certificate (definitely over 50k a year, probably more like 60k). It depends what your location is a lot, for both your pay and your odds. But it’s a very small field and not many people are swarming into it. It’s probably not the worst of ideas to go for it.

careerbuilder has a great salary calculator.

## How to enter in log base 3 to a calculator?

I have the TI-84 Plus.
I would like to enter in log_3, or „log base 3.“
Do I need to change the equation to do this, or is there a way to enter it in?
My equation is log_3(1-x)+2.
Thank you!
My equation is:

f(x) = log (base/subscript 3) (1-x)+2.
I think that would be correct, Ray. Thanks for asking!

I think you will probably need to use the change of base formula for this.

This says log (base 3) A, where A is a number, is equal to [log (base 10) A] / [log (base 10) 3].

There is something odd about your „equation“ – it does not contain an equals sign.

Even as an expression there is something odd about it.

Is it [log (base 3) (1 – x)] + 2 ?