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3 years ago

PLEASE HELP ME FILL THIS CHART OUT I WILL MARK BRAINIEST

Mathematics

1 answer:

svetlana [45]3 years ago

8 0

Answer:

volume of can #1: 56.55 SA: 95.25

volume of can #2: 113.1 SA: 138.23

volume of can #3: 113.1 SA: 50.3

volune of can #4: 75.4. SA: 113.1

I think i did it right

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debt payments of $800 due now and $1400 due in 5 months are to be repaid by a payment of $1000 in 3 months and a final payment i

loris [4]

Answer:

$1,304.70

Step-by-step explanation:

If interest 6% annually, monthly is 0.5%.

The debt in 5 months will be 800 plus compounded interest for 5 months plus new due debt

$800(1.005)^5+1400=2220.201$

In 3 more months the debt will be 2220.201 plus compounded interest for 3 months minus payment

$2220.201(1.005)^3-1000=1253.67$

After 8 months the debt would be 1253.67 plus compounded interest for 8 months

$2220.201(1.005)^3=1304.70$

Then the size of the final payment would be $1,304.70

5 0

3 years ago

If the ratio of men to women is 3:2 and there are 120 men, how many women would there be?

zavuch27 [327]

There would be 80 women.
Here's the work:
120 divided by 3 = 40
If the ratio is 3:2, there is always one less woman then men, so you subtract the 40 men in a group and you get your answer of 80

Hope I helped!! :)

5 0

2 years ago

Read 2 more answers

Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area: where V = volume (mm3

Alex

Answer:

V = 20.2969 mm^3 @ t = 10

r = 1.692 mm @ t = 10

Step-by-step explanation:

The solution to the first order ordinary differential equation:

$\frac{dV}{dt} = -kA$

Using Euler's method

$\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i} \\$

Where initial droplet volume is:

$V(0) = \frac{4pi}{3} * r(0)^3 = \frac{4pi}{3} * 2.5^3 = 65.45 mm^3$

Hence, the iterative solution will be as next:

i = 1, ti = 0, Vi = 65.45

$V'_{i} = -k *4pi*(\frac{3*65.45}{4pi})^(2/3) = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88$

i = 2, ti = 0.5, Vi = 63.88

$V'_{i} = -k *4pi*(\frac{3*63.88}{4pi})^(2/3) = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33$

i = 3, ti = 1, Vi = 62.33

$V'_{i} = -k *4pi*(\frac{3*62.33}{4pi})^(2/3) = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813$

We compute the next iterations in MATLAB (see attachment)

Volume @ t = 10 is = 20.2969

The droplet radius at t=10 mins

$r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\$

The average change of droplet radius with time is:

Δr/Δt = $\frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min$

The value of the evaporation rate is close the value of k = 0.08 mm/min

Hence, the results are accurate and consistent!

5 0

3 years ago

A basketball player gets 2 free-throw shots when she is fouled by a player on the opposing team. She misses the first shot 40% o

yan [13]

<span>The probability of missing the first shot is 40%. When she misses the first shot, the probability of missing the second shot is very, very low, only 5%. That means that the probability of missing both shots must be much smaller than the probability of missing the first shot.

40/100 * 5/100
0.4 * 0.05
= 0.02
= 2%</span>

5 0

3 years ago

What kind of background would fit?

Dafna11 [192]

Answer:

Is this even math

Step-by-step explanation:

No it isn’t

Just ask a forum or something and not an education site lol

6 0

2 years ago