Help me with this!!!!

if 5.25 MB have been downloaded and 2.25 MB are left. what percent of the application has already been downloaded?

ollegr [7]

5.25 + 2.25 = 7.50 MB

x/100 x (7.50) = 5.25
7.50x/100 = 5.25
7.50x = 5.25(100)
7.50x = 525
x = 525/7.50
x = 70
Hence, 70% of the application has already been downloaded.

6 0

4 years ago

Maggie needs to spend at least six hours each week practicing the piano. She has already practiced 3

grandymaker [24]

Answer:

t ≥ 1.5

Step-by-step explanation:

Lets take time for practicing the piano to be t

The number of hours to practice per week should be at least 6 hours

This is written as t ≥ 6

She already practiced 3 hours this week, thus the remaining hours to practice should be

t ≥ 3

The minimum remaining hours to practice for the remaining 2 days should be

t=3

If these hours are evenly divided, then in a single day she should practice

for t ≥ 1.5

8 0

3 years ago

Hey I’m struggling with number 13 so please show your work. Thx

Dafna11 [192]

Answer:

$\frac{5}{12}$

Step-by-step explanation:

Because $\frac{3}{5}$ is being multiplied by S, we can divide $\frac{3}{5}$ from both sides of the equation. This will give us:

$\frac{1}{4}$ ÷ $\frac{3}{5}$

But, that looks a bit hectic. Instead of dividing, you can multiply by the reciprocal (which is essentially how to divide fractions). So, instead of $\frac{1}{4}$ ÷ $\frac{3}{5}$ , you get:

$\frac{1}{4}$ × $\frac{5}{3}$

When multiplying fractions, remember you can just multiply straight across-- numerator x numerator, then denominator x denominator.

By doing that, you get the fraction $\frac{5}{12}$

$\frac{5}{12}$ cannot be simplified any more, so S= $\frac{5}{12}$ is your answer :)

I hope this helps

6 0

3 years ago

If f(z)= sqrt z2+8z , then how much does f(z) increase as z goes from 7 to 8?

Nuetrik [128]

Answer:

B

Step-by-step explanation:

$f(7)=\sqrt{7^2+8*7}=\sqrt{49+56}=\sqrt{105} \approx 10.247\\f(8)=\sqrt{8^2+8*8}=\sqrt{128}=8\sqrt{2} \approx 11.314\\f(8)-f(7)=11.314-10.247=1.067 \approx 1.07$

3 0

3 years ago

A fair die is rolled 12 times. the number of times an even number occurs on the 12 rolls has

bonufazy [111]

Answer:

Step-by-step explanation:

For a fair die, there are six likely options; 1, 2, 3, 4, 5, and 6

the probability of a even number is 3/6 = 0.5

Since the results of the die roll is independent and each trial is mutually exclusive, the distribution to explain the probability of occurrence will follow a binomial distribution such that n is the number of trials

x = number of successful throws

therefore for a Binomial distribution where

P(X =x) = nCx . P^x . (1-P)^ (n-x)

since p = 0.5, and n = 12, the distribution follows

P(X = x) = 12Cx . 0.5^x . (1 - 0.5)^(12- x)

= 12Cx . 0.5^x . 0.5)^(12- x)

where x = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)

since we are interested in the probability of the number of times an even number occurs

it can occur either as P(X = 0), P(X =1), P(X =2), P(X =3), P(X =4), P(X =5), P(X =6), P(X =7), P(X =8), P(X =9), P(X =10), P(X =11), and P(X =12)

For no even number in 12 rolls,

P(X = 0) = 12C0 . 0.5^0 . 0.5^(12- 0) = 0.000244

For one even number in 12 rolls,

P(X = 1) = 12C1 . 0.5^1 . 0.5^(12- 1) = 0.002930

For two even number in 12 rolls,

P(X = 2) = 12C2 . 0.5^2 . 0.5^(12- 2) = 0.016113

For three even number in 12 rolls,

P(X = 3) = 12C3 . 0.5^3 . 0.5^(12- 3) = 0.053711

For four even number in 12 rolls,

P(X = 4) = 12C4 . 0.5^4 . 0.5^(12- 4) = 0.120850

For five even number in 12 rolls,

P(X = 5) = 12C5 . 0.5^5 . 0.5^(12- 5) = 0.193359

For six even number in 12 rolls,

P(X = 6) = 12C6 . 0.5^6 . 0.5^(12- 6) = 0.225586

For seven even number in 12 rolls,

P(X = 7) = 12C7 . 0.5^7 . 0.5^(12- 7) = 0.193359

For eight even number in 12 rolls,

P(X = 8) = 12C8 . 0.5^8 . 0.5^(12- 8) = 0.120850

For nine even number in 12 rolls,

P(X = 9) = 12C9 . 0.5^9 . 0.5^(12- 9) = 0.053711

For ten even number in 12 rolls,

P(X = 10) = 12C10 . 0.5^10 . 0.5^(12- 10) = 0.016113

For eleven even number in 12 rolls,

P(X = 11) = 12C11 . 0.5^11 . 0.5^(12- 11) = 0.002930

For twelve even number in 12 rolls,

P(X = 12) = 12C12 . 0.5^12 . 0.5^(12- 12) = 0.000244

Final test summation[P(X)] = 1

i.e.

P(X = 0) + P(X =1) + P(X =2) + P(X =3) + P(X =4) + P(X =5) + P(X =6) + P(X =7) + P(X =8) + P(X =9) + P(X =10) + P(X =11) + P(X =12) = 1

Hence since 0.000244 + 0.002930 + 0.016113 + 0.053711 + 0.120850 + 0.193359 + 0.225586 + 0.193359 + 0.120850 + 0.053711 + 0.016113 + 0.002930 + 0.000244 = 1.000000,

the probability value stands

7 0

3 years ago