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Ber [7]
2 years ago
7

I neeed heeelppppppppp

Mathematics
1 answer:
zheka24 [161]2 years ago
4 0

Answer:

P(x) =-2x^{2} + 304x  - 8510

x = 37 smallest B.E.P

Step-by-step explanation:

P = R - E

P =-2x^{2} + 317 x

E =  13x + 8510

P =-2x^{2} + 317 x - 13x - 8510

P(x) =-2x^{2} + 304x  - 8510

B.E.P.(X) where  R=E

-2x^{2} + 304 x  -  8510 = 0

2(x-37)(x-115)

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3/4•3/4???? I can’t do it in my head and I don’t know how to make a decimal into a fraction
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Answer:

0.5625

Step-by-step explanation:

3/4 as a decimal is 0.75

0.75×0.75=0.5625

4 0
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a class conduct a survey of 1000 students At one point 6% of the students forgot their locker combination, how many forgot their
jek_recluse [69]
You need to multiply 1000 by 0.06 (6% in decimal form) to get the answer. When you multiply you get 60 so the answer is 60 students forgot their locker combination.
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3 years ago
Helppp me plsssssssss<br><br>​
Oliga [24]

Answer:

The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

  • \sf \:\:\:\:\:\:\:\:\:\:x_{k}=35
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k}=50
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k-1}=34
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k+1}=42
  • \sf \:\:\:\:\:\:\:\:\:\:h=5

{\bf \:\: {By\:using\:the\: formula}} \\ \\

\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}}} \\ \\

\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 \times \dfrac{(50 - 34)}{ ( 2 \times 50 - 34 - 42)}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 \times \dfrac{16}{24}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+\dfrac{10}{3}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\

\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\

{\large{\frak{\pmb{\underline{Additional\: information }}}}}

MODE

  • Most precisely, mode is that value of the variable at which the concentration of the data is maximum.

MODAL CLASS

  • In a frequency distribution the class having maximum frequency is called the modal class.

{\bf{\underline{Formula\:for\: calculating\:mode:}}} \\

{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}} \\ \\

Where,

\sf \small\pink{ \bigstar} \: x_{k}= lower\:limit\:of\:the\:modal\:class\:interval.

\small \blue{ \bigstar}\sf \: f_{k}=frequency\:of\:the\:modal\:class

\sf \small\orange{ \bigstar}\: f_{k-1}=frequency\:of\:the\:class\: preceding\:the\;modal\:class

\sf \small\green{ \bigstar}\: f_{k+1}=frequency\:of\:the\:class\: succeeding\:the\;modal\:class

\small \purple{ \bigstar}\sf \: h= width \:of\:the\:class\:interval

7 0
3 years ago
B is the midpoint of ac. Ab = x+9 and bc = 3x-7 find x and ac
natali 33 [55]

To solve this problem, we need to know 2 relationships:

<h2>1. AC = AB + BC</h2>

The distance of AC is the sum of AB and BC.

AC = AB + BC

We know this since the distance of going from A to C (AC) is the same as going from A to B (AB), then B to C (BC).

<h2>2. AB = BC</h2>

The distance of AB is the same as AC.

AB = BC

We know this since B is in the middle of AC, so the distance from B to A (BA) is the same as the distance from B to C (BC).

You can see the attached image (at the bottom) for a visualization of this.

<h2>Putting them together</h2>

Since we know the values of AB and BC...

AB = x+9\\BC = 3x-7

...we can put these values into our 2nd equation and solve for x:

AB = BC\\x + 9 = 3x -7

Add 7 to both sides:

x + 16 = 3x

Subtract x from both sides:

16 = 2x

Divide both sides by 2:

8 = x\\x = 8

Knowing x, we can find the distance of AC using our first equation.

AC = AB + BC

Let's put in the values of AB and BC:

AC = (x+9) + (3x-7)

Before we put in x = 8, we can simplify this:

AC = (x+9) + (3x-7)\\AC = x + 9 + 3x -7\\AC = x + 3x + 9 -7\\AC = 4x + 9 - 7\\AC = 4x+2

We group x and 3x and add those together. Then we subtract 7 from 9.

With this equation, we can put in x = 8:

AC = 4x +2\\AC = 4*8 + 2

Since 4 * 8 = 32:

AC = 4 * 8 + 2\\AC = 32 + 2\\AC = 34

Finally, we have found both x and AC.

<h2>Answer</h2>

x = 8

AC = 34

7 0
3 years ago
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