5 horses with a combined weight of 6,240 pounds
horse trailer has a limit of 8,000 pounds of horse and tacks.
Find the weight of tack, t, for each horse that Cari can allow.
5t + 6,240 < 8,000
5t + 6,240 < 8,000
- 6,240 -6,240
5t < 1,760
<span>÷5 ÷5 </span>
t < 352
Each horse can carry a tack that weighs no more than 352 pounds.
Any tack that weighs beyond <span>352 </span>pounds is will exceed the maximum carrying capacity of the horse trailer.
1) You must add 4 to each side to complete the square.
2) You must add 16 to each side to complete the square.
3) You must add 27 to each side to complete the square.
Explanation:
1) x²-4x=0
To find the number that we add to both sides, we look at b, the cofficient of x. It is -4. We divide it by 2 and square it; -4/2 = -2; (-2)² = 4. This is the value that we add to both sides.
2) x²-8x=6
-8/2 = -4; (-4)²=16
We add 16 to each side to complete the square.
3) 3x²+18x=24
First we can factor a 3 out of the left side:
3(x²+6x) = 24
Our b value is now 6. 6/2 = 3; 3²=9. The 9 would, however, go in the parentheses, so it would be multiplied by 3, which makes 27; this means we would add 27 to both sides.
Ok so we are using the Pythagorean theorem so were gonna do A^2 + B^2 = C^2
It don't say which leg we have so you can use either a or b.
Which would get us 14^2 + B^2 = 50^2
14*14 =196 , 50 *50 =2500
Then we have to subtract the two.
So 2500 - 196 = 2,304
Then you take the take the square root of that answer and you have your other leg.
The square root of 2,304 is 48
So the missing leg = 48
Answer:
Step-by-step explanation:
Given that,
The radius of a cylinder, r = 5 cm
Height of the cylinder, h = 5 cm
We need to find the lateral surface area of the cylinder. The formula for the lateral surface area of the cylinder is given by :
Put all the values,
So, the lateral surface area of the cylinder is 
.
Answer:
a) 229 and 305 days
b) 229 days or less
c) 305 days or more
Step-by-step explanation:
The Empirical Rule(68-95-99.7 rule) states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 267
Standard deviation = 19
(a) Between what values do the lengths of the middle 95% of all pregnancies fall?_____________and___________days
By the Empirical rule, 95% of all pregnancies fall within 2 standard deviations of the mean.
So
267 - 2*19 = 229 days
to
267 + 2*19 = 305 days
(b) How short are the shortest 2.5% of all pregnancies?______days or less
95% of all pregnancies fall within 2 standard deviations of the mean. The other 5% are more than 2 standard deviations from the mean. Since the distribution is symmetric, 2.5% is more than 2 standard deviations below the mean(shortest 2.5%) and 2.5% is more than 2 standard deviations above the mean(longest 2.5%). So
267 - 2*19 = 229 days
c) How long do the longest 2.5% of pregnancies last?________days or more
Explanation in b)
267 + 2*19 = 305 days
