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Vlad [161]
3 years ago
13

How do I solve 12 < 4 + 2p

Mathematics
2 answers:
pickupchik [31]3 years ago
7 0

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

                Hello!

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

  • First I would flip the inequality too: 2p + 4 > 12
  • Then you want to subtract four (since we have to use the inverse operation) from both sides:

2p + 4 > 12

      -4    -4

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

2p > 8

  • Now, we divide both sides by 2:

2p > 8

-----------

2       2

  • The final result is p > 4

~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ

blondinia [14]3 years ago
6 0

Answer:

p > 4

Step-by-step explanation:

-2p < 4 - 12

Calculate^

-2 p < -8

Divide Both sides by -2^

And your answer is p > 4

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Jerome burns 4 cal/min walking and 10 cal/min running. He walks between 10 and 20 min each day and runs between 30 and 45 min ea
erik [133]

Answer:

20 minutes should spend on walking and 30 minutes on running.

Step-by-step explanation:

Let x represents the minutes spent on walking and y represents the minutes spent on running,

∵ He burns 4 cal/min walking and 10 cal/min running,

So, the total calories burnt,

Z = 4x + 10y

Which has to be maximised.

Now, he walks between 10 and 20 min each day,

i.e. 10 < x < 20,

Also, he runs between 30 and 45 min each day,

i.e. 30 <  y < 45,

By graphing 10 < x < 20 and 30 <  y < 45,

We obtained the vertices of feasible region,

(10, 45), (20, 45), (10, 30) and (20, 30)

At (10, 45),

Z = 4(10) + 10(45) = 40 + 450 = 490,

At (20, 45),

Z = 4(20) + 10(45) = 80 + 450 = 530,

At (10, 30),

Z = 4(10) + 10(30) = 40 + 300 = 340,

At (20, 30)

Z = 4(20) + 10(30) = 80 + 300 = 380

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5 0
3 years ago
Quick pls
navik [9.2K]

First, let's see how 23 compares with the squares of the positive whole numbers on the number line.

1² = 1

2² = 4

3² = 9

4² = 16

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The value of 23 is right between the square of 4 and the square of 5. Thus, the value √23 will be between 4 and 5.

Since 23 is much, much closer to the square of 5 than the square of 4, we can assume that the value √23 will be closer to 5 on the number line than 4.

Look at the attached image to see where I plotted the approximate location of √23.

You will realize that this approximation is pretty close since the actual value is roughly 4.80.

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7 0
3 years ago
The computers of nine engineers at a certain company are to be replaced. Four of the engineers have selected laptops and the oth
Gala2k [10]

Answer:

(a) There are 70 different ways set up 4 computers out of 8.

(b) The probability that exactly three of the selected computers are desktops is 0.305.

(c) The probability that at least three of the selected computers are desktops is 0.401.

Step-by-step explanation:

Of the 9 new computers 4 are laptops and 5 are desktop.

Let X = a laptop is selected and Y = a desktop is selected.

The probability of selecting a laptop is = P(Laptop) = p_{X} = \frac{4}{9}

The probability of selecting a desktop is = P(Desktop) = p_{Y} = \frac{5}{9}

Then both X and Y follows Binomial distribution.

X\sim Bin(9, \frac{4}{9})\\ Y\sim Bin(9, \frac{5}{9})

The probability function of a binomial distribution is:

P(U=k)={n\choose k}\times(p)^{k}\times (1-p)^{n-k}

(a)

Combination is used to determine the number of ways to select <em>k</em> objects from <em>n</em> distinct objects without replacement.

It is denotes as: {n\choose k}=\frac{n!}{k!(n-k)!}

In this case 4 computers are to selected of 8 to be set up. Since there cannot be replacement, i.e. we cannot set up one computer twice or thrice, use combinations to determine the number of ways to set up 4 computers of 8.

The number of ways to set up 4 computers of 8 is:

{8\choose 4}=\frac{8!}{4!(8-4)!}\\=\frac{8!}{4!\times 4!} \\=70

Thus, there are 70 different ways set up 4 computers out of 8.

(b)

It is provided that 4 computers are randomly selected.

Compute the probability that exactly 3 of the 4 computers selected are desktops as follows:

P(Y=3)={4\choose 3}\times(\frac{5}{9})^{3}\times (1-\frac{5}{9})^{4-3}\\=4\times\frac{125}{729}\times\frac{4}{9}\\  =0.304832\\\approx0.305

Thus, the probability that exactly three of the selected computers are desktops is 0.305.

(c)

Compute the probability that of the 4 computers selected at least 3 are desktops as follows:

P(Y\geq 3)=1-P(Y

Thus, the probability that at least three of the selected computers are desktops is 0.401.

6 0
3 years ago
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