Find the number that makes the ratio equivalent to 9:1 Answer has to end with :8

Find mABC A-54 B-135 C-15 D-126

bagirrra123 [75]

It’s B

Glad I could help

7 0

3 years ago

Median and mean math questions

Harlamova29_29 [7]

Answer: #7 D) neither set. #6. C) the sum of the data values must be 75. #5. The 6th missing temperature must be 80 degrees

Step-by-step explanation: #7. if you eliminate 3 numbers on each side on set A, then the numbers remaining are 2 and 19. In order to find the median add the two numbers, 19+2 which equals 21. Then divide that number by 2 to find your median, or in other words find the average, 21 divided by 2 is 10.5 which is not 12.5, so that set doesn’t work. In set B eliminate 2 numbers on each side. That will leave you with 10 and 9, then again find the average of the numbers. 10+9=19, and 19 divided by 2= 9.5. So therefore neither of the sets have a median of 12.5

#6. If there are five numbers in the data set and you are finding the average to be 15, then the five values must equal 75. In order to find the average of five numbers you must first add all of your values, and then divide by five, for there are five numbers. If you add all of your numbers up it must total 75, because 75 divided by 5= 15 which is the mean. Any other total would not work.

#5. The median of the original set is 72 degrees. In order to find the missing number you need to do some check and guess. Take the original median and add a number relatively close to the other numbers, for example 75, then add the two numbers together, 72+75=147. Then divide that by two, 147 divided by 2= 73.5 which is not the correct median number, its too low, so experiment with higher numbers and repeat the same process until you find a number that, if added to 72, totaled, and then divided by 2 equals 76.

I really hope this helped :)

7 0

3 years ago

Find the derivative.

Aleksandr [31]

Answer:

Using either method, we obtain: $t^\frac{3}{8}$

Step-by-step explanation:

a) By evaluating the integral:

$\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du$

The integral itself can be evaluated by writing the root and exponent of the variable u as: $\sqrt[8]{u^3} =u^{\frac{3}{8}$

Then, an antiderivative of this is: $\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}$

which evaluated between the limits of integration gives:

$\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}$

and now the derivative of this expression with respect to "t" is:

$\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}$

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

$g(x)=\int\limits^x_a {f(t)} \, dt$

is continuous on [a,b], differentiable on (a,b) and $g'(x)=f(x)$

Since this this function $u^{\frac{3}{8}$ is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means: