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Irina18 [472]
3 years ago
13

What is the real number value of m^3+ √12 m when 5m^2 = 45 ?

Mathematics
1 answer:
babymother [125]3 years ago
5 0
<h3>Answer: H. 33</h3>

=============================================

Work Shown:

Solve 5m^2 = 45 for m to get

5m^2 = 45

m^2 = 45/5

m^2 = 9

m = sqrt(9)

m = 3

I'm making m to be positive so that way the expression 12m is not negative. Otherwise, sqrt(12m) would not be a real number result.

--------------

Plug m = 3 into the expression we want to evaluate

m^3 + sqrt(12m)

3^3 + sqrt(12*3)

27 + sqrt(36)

27 + 6

33

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Nuts Berry Farm sells nuts for $60 for 4 pounds. Today, and just for today only,
tino4ka555 [31]

Answer:

$10.50

Step-by-step explanation:

First, find the cost of a pound of nuts.

Divide the price by the number of pounds:

60/4

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15(0.7)

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7 0
3 years ago
Can you help me with number 6? <br> Confused abit <br> Please
Sunny_sXe [5.5K]

You can see the three diagram attached. Each link is labeled with the probability: you have probability 1/6 that a six is rolled, and 5/6 that it is not rolled.


To answer the questions, find the path that brings you to the desired outcome, and multiply all the labels you meet.


First question:

To get three sixes, you have to choose the left path at each roll. The probability is always 1/6, so the answer is


\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{6^3}


Second question:

To get no sixes, you have to choose the right path at each roll. The probability is always 5/6, so the answer is


\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^3}{6^3}


Third question:

To get exactly one six, it can either be the first, second or third roll.


In all cases, you have to choose the left path once and the right path twice: left-right-right mean that you get the six in the first roll, right-left-right means that you get the six in the second roll, right-right-left means that you get the six in the third roll.


In every case, the left turn has probability 1/6, and the right turn has probability 5/6. The probability of each combination is thus


\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^2}{6^3}


And since there are three of these combinations, The answer is


3\frac{5^2}{6^3}


Fourth question:

Since the question suggests to use what we already achieved, let's do it: having at least one six is the complementary event of having no sixes at all. If an event has probability p, its complementary has probability 1-p. So, since the probability of no sixes is known, the probability of at least one six is


1 - \frac{5^3}{6^3}

4 0
3 years ago
whole milk is 4% butterfat how much skim milk with 0% butterfat should be added to 32 ounces of whole milk to obtain a mixture t
vivado [14]

Answer:

whole milk 4 ---------------- 32 ounces

skim milk 0 ---------------- x ounces

Mixture 2.5 ---------------- 32 + x ounces

 

4*32+0x =2.5(32 +x)

128+0x= 80+2.5 x

0 x + -2.5 x = 80 - -128

-2.5 x = -48

/ -2.5

x = 19.2

19.2 ounces of 0 % fat skim milk

4 0
3 years ago
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