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3 years ago

12

HELP PLEASE WILL GIVE BRAINLIST IF CORRECT

1 answer:

Nat2105 [25]3 years ago

4 0

Answer:

a.) First check on the height of the bar that ranges between 20 and 24 which is 6. The 6 is the number of students who bought 20 books or more, so get the percentage of those students who bought 20 or more books; divide 6 which is the number of students who bought 20 books or more with the total number of students which is 40 then multiply with 100.

% of students who bought 20 or more books = 6/40 ×100

=15

Step-by-step explanation:

b.) Find the area of a trapezium

= 1/2 × {a+b} × h

= 1/2 × {9 + 13} ×12

= 132

= 1/2 × {13 + 6} × 12

= 108

Mean no. of books bought = 132 + 108

= 240

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PLEASE GET THE FILL IN BLANK ANSWER RIGHT ixl pLEASE

jolli1 [7]

Required equation in slope intercept form is: $y=5x+4$

Step-by-step explanation:

Given

Let m denote the slope of line

m = 5

the line passes though (1,9)

Slope-intercept form of equation of line is:

$y= mx+b$

Putting m=5 in the equation

$y = 5x+b$

Putting (1,9) in the equation

$9 = 5(1) +b\\9 = 5+b\\b = 9-5\\b = 4$

Putting the value of b

$y = 5x+4$

Hence,

Required equation in slope intercept form is: $y=5x+4$

Keywords: Equation, linear equation

Learn more about equation of line at:

#LearnwithBrainly

3 0

3 years ago

What percent is represented by the diagram

Firdavs [7]

Answer:

180%

Step-by-step explanation:

A full box shaded in plus 80% of another box is 180%.

4 0

3 years ago

Read 2 more answers

The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

We can use the sample about 42 days.

7 0

3 years ago

The sum of 5 consecutive numbers is 200. What is the smallest number?

Vikentia [17]

Answer:

Hey!

Your answer is 44!

Step-by-step explanation:

Here's how...

x+(x+2)+(x+4)+(x+6)+(x+8)=200

5x+20=200 (minus 20 on both sides to get 5x by itself)

5x=180 (divide both sides by 5 to get just x)

x=36

5th number - x+8=36+8=44 (simple substitution and addition)

I HOPE THIS HELPS!

6 0

3 years ago

Read 2 more answers

Evaluate 10 m +\frac {n^2}410m+ 4 n 2 10, m, plus, start fraction, n, start superscript, 2, end superscript, divided by, 4, en

DerKrebs [107]

Evaluate the expression $10m + \frac{n^2}{4}$ where m=5 and n=4

We replace the value of m and n to evaluate

m=5 and n=4

$10m + \frac{n^2}{4}$

$10(5) + \frac{4^2}{4}$

$50 + \frac{16}{4}$

50 + 4 = 54

The value of the given expression is 54 when m=5 and n= 4

5 0

3 years ago