What’s the expression equivalent to

Mathematics

1 answer:

Natasha_Volkova [10]4 years ago

5 0

the answer is A (need more characters to add this so here)

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What is the lateral area of a right cylinder if the radius of the base is 12 mm and the altitude of the cylinder is 5 times the

schepotkina [342]

Lateral Area of a cylinder is ; 2πR.H

R=12 mm and H= 15 mm

Then LATERAL area = 2π(12)(15) =48π ≈ 150 mm²

5 0

3 years ago

Solve 2x + 3y = 16 4x+10y = 40

Aleksandr-060686 [28]

Answer:

x = 5, y = 2

Step-by-step explanation:

First, find the value of x in respect to y:

2x + 3y = 16
2x = 16 - 3y

x = 8 - 3/2y

Then, substitute this expression into the second equation (Since we've established x is equal to 8 - 3/2y, we can put that in place of x):

4(8 - 3/2y) + 10y = 40

Then, distribute and solve for y:

32 - 6y + 10y = 40
32 + 4y = 40
4y = 8
y = 2

Now that we know what y equals, we can substitute the value of y in the equation to find x:
2x + 3(2) = 16
2x + 6 = 16
2x = 10

x = 5

8 0

3 years ago

A gym charges an initial fee plus $25 per month. John paid $200 for six months. Assume the relationship is linear.

charle [14.2K]

Step-by-step explanation:

a. The rate of change is $33.33 per month

200/6 = approximately $33.33 per month

This means that in addition to the initial fee, Josh has to pay $33.33 per month for the gym membership.

b. The initial value is $25. This is a constant value and a one time fee to open and attain a gym membership.

7 0

3 years ago

The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approxi

bazaltina [42]

Answer:

$425.6 should be budgeted for weekly repairs and maintenance.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean $400 and standard deviation $20.

This means that $\mu = 400, \sigma = 20$

How much should be budgeted for weekly repairs and maintenance to provide that the probability the budgeted amount will be exceeded in a given week is only 0.1?

This is the 100 - 10 = 90th percentile, which is X when Z has a p-value of 0.9, so X when Z = 1.28.

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 400}{20}$