All categories

2 years ago

PLS HELP I WILL GIVE BRIANLY!

Mathematics

1 answer:

sertanlavr [38]2 years ago

7 0

What was the discount? There isn’t enough information for me to answer it

You might be interested in

I’m not sure why the picture didn’t work

lutik1710 [3]

Answer:

4.1

Step-by-step explanation:

6 0

2 years ago

Helppp me i neeed help

dexar [7]

Total no. of buttons = 300 + 700 + 1000

+ 500 = 2500

Possibility for a person to pick a pink button = 300/2500 = (3/25)%

Amount of pink buttons she can expect her friends to pick = 50(3/25) = 6

She can expect to make 6 pink barrettes.

6 0

2 years ago

Read 2 more answers

A number line ranging from 5 to 45 with a box encompassing the numbers 20, 25, and 30.

irina1246 [14]

Answer:

the maximum value is 45

Step-by-step explanation:

A box of leaf plot has 5 points and they are:

The minimum, the first quartile (Q1) the median (Q2), the third quartile (Q3), and the maximum. The box encompasses Q1, Q2, and Q3. The maximum is after Q3 and the minimum is before Q1.

In this case,

Q1, Q2, and Q3 are in a range 20-30, thus the maximum come after 30, which is 45

4 0

1 year ago

A jewelry supply shop buys silver chains from a manufacturer for c dollars each, and then sells the chains at a 57% markup. Find

g100num [7]

I believe it is 70.65

3 0

2 years ago

Read 2 more answers

After the mill in a small town closed down in 1970, the population of that town started decreasing according to the law of expon

wlad13 [49]

Answer:

$P_o = \frac{143000}{e^{-20*0.01303024661}}=110193.69$

And we can round this to the nearest up integer and we got 110194.

Step-by-step explanation:

The natural growth and decay model is given by:

$\frac{dP}{dt}=kP$ (1)

Where P represent the population and t the time in years since 1970.

If we integrate both sides from equation (1) we got:

$\int \frac{dP}{P} =\int kdt$

$ln|P| =kt +c$

And if we apply exponentials on both sides we got:

$P= e^{kt} e^k$

And we can assume $e^k = P_o$

And we have this model:

$P(t) = P_o e^{kt}$

And for this case we want to find $P_o$

By 1990 we have t=20 years since 1970 and we have this equation:

$143000 = P_o e^{20k}$

And we can solve for $P_o$ like this:

$P_o = \frac{143000}{e^{20k}}$ (1)

By 2019 we have 49 years since 1970 the equation is given by:

$98000 = P_o e^{49k}$ (2)

And replacing $P_o$ from equation (1) we got:

$98000=\frac{143000}{e^{20k}} e^{49k} =143000 e^{29k}$

We can divide both sides by 143000 we got:

$\frac{98000}{143000} =0.685 = e^{29k}$

And if we apply ln on both sides we got:

$ln(0.685) = 29k$

And then k =-0.01303024661[/tex]

And replacing into equation (1) we got:

$P_o = \frac{143000}{e^{-20*0.01303024661}}=110193.69$

And we can round this to the nearest up integer and we got 110194.

7 0

2 years ago