HELP ASAP!!! Due at 11:59

10×(6×5) +9^3 ×8<br><img src="https://tex.z-dn.net/?f=10%20%5Ctimes%20%286%20%5Ctimes%205%29%20%2B%20%20%7B9%7D%5E%7B3%7D%20%20%

Minchanka [31]

Answer: 6132

Step-by-step explanation:

10x(6x5)+9^3 x8

10x30+9^3 x8

10x30+729x8

300+729x8

300+5832

6132

Hope this helps!

3 0

3 years ago

Three times the difference between 4 times h and five is 45

Anestetic [448]

3(4h-5) =45

12h-15=45

+15 +15

12h = 60

÷12 ÷12

h = 5

5 0

3 years ago

Explain why (–4x)0 = 1, but –4x0 = –4.

hodyreva [135]

Any number raised to power of 0 is 1.
In (-4x)^0, because of the parenthesis, the whole number is being raided to 0, which gives 1.
But in -4x^0, only x is being raised to 0, thus -4x^0 = -4(1) = -4.

5 0

3 years ago

An experiment consists of choosing objects without regards to order. Determine the size of the sample space when you choose the

sdas [7]

Answer:

a

$n= 75, 582$

b

$n= 2300$

c

$n = 253$

Step-by-step explanation:

Generally the size of the sample sample space is mathematically represented as

$n = \left N } \atop {}} \right. C_r$

Where N is the total number of objects available and r is the number of objects to be selected

So for a, where N = 19 and r = 8

$n = \left 19 } \atop {}} \right. C_8 = \frac{19 !}{(19 - 8 )! 8!}$

$= \frac{19 *18 *17 *16 *15 *14 *13 *12 *11! }{11 ! \ 8!}$

$n= 75, 582$

For b Where N = 25 and r = 3

$n = \left 25 } \atop {}} \right. C_3 = \frac{25 !}{(19 - 3 )! 3!}$

$= \frac{25 *24 *23 *22 ! }{22 ! \ 3!}$

$n= 2300$

For c Where N = 23 and r = 2

$n = \left 23 } \atop {}} \right. C_2 = \frac{23 !}{(23 - 2 )! 2!}$

$= \frac{23 *22 *21! }{21 ! \ 3!}$

$n = 253$

4 0

3 years ago

A veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses

Licemer1 [7]

Answer:

Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

Step-by-step explanation:

We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.

So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = average age of the random sample of horses with colic = 12 yrs

$\mu$ = average age of all horses seen at the veterinary clinic = 10 yrs

$\sigma$ = standard deviation of all horses coming to the veterinary clinic = 8 yrs

n = sample of horses = 60

So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P( $\bar X$ $\geq$ 12)

P( $\bar X$ $\geq$ 12) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{12-10}{\frac{8}{\sqrt{60} } }$ ) = P(Z $\geq$ 1.94) = 1 - P(Z < 1.94)

= 1 - 0.97381 = 0.0262

Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

4 0

3 years ago