-28/((-12+9)-(9+12/3)+1)

If a/4 = b/7, what is the value of (a-b) / b?

DerKrebs [107]

First we will find out the values of a
a/4=b/7
a=4b/7

Now
(a-b)/b
{4b/7-b}/b
(4b-7b)/7b
-3b/7b
-3/7

8 0

2 years ago

Read 2 more answers

Can someone help me please and thanks .

enyata [817]

Answer:

$180.

Step-by-step explanation:

First lets find the interest rate.

6.1 ÷ 100 = 0.061

0.061 x $200 = 12.2

based on the calculations, 12.2 is our interest rate in a year. Now lets find the interest rate in 15 years.

12.2 x 15 = $183

So $183 is our precise interest rate in 15 years, but lets round the interest rate to the nearest 10 dollars. Therefore, $180 is our answer.

6 0

2 years ago

12% of children are nearsighted, but this condition often is not detected until they go to kindergarten. A school district tests

Klio2033 [76]

Answer:

There is a 34.60% probability that 0 or 1 of them is nearsighted.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they are nearsighted, or they are not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

12% of children are nearsighted. This means that $p = 0.12$ .

A school district tests all incoming kindergarteners' vision. In a class of 18 kindergarten students, what is the probability that 0 or 1 of them is nearsighted?

There are 18 students, so $n = 18$

This probability is:

$P = P(X = 0) + P(X = 1)$

So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{18,0}.(0.12)^{0}.(0.88)^{18} = 0.1002$

$P(X = 1) = C_{18,1}.(0.12)^{1}.(0.88)^{17} = 0.2458$

So

$P = P(X = 0) + P(X = 1) = 0.1002 + 0.2458 = 0.3460$

There is a 34.60% probability that 0 or 1 of them is nearsighted.

5 0

3 years ago

7.2÷(6.6×1.1+(2.2) to the second power-3.1)

Hoochie [10]

Answer:

0.98001931 is your answer : >

6 0

3 years ago

Read 2 more answers

Solve the proportion: 20/16 = x/80 question 19 options: 6.25 1600 100 160

Aloiza [94]

$\dfrac{20}{16}=\dfrac{20:4}{16:4}=\dfrac{5}{4}\\\\\dfrac{5}{4}=\dfrac{x}{80}\ \ \ \ |cross\ multiply\\\\4x=5\cdot80\\\\4x=400\ \ \ \ |divide\ both\ sides\ by\ 4\\\\\boxed{x=100}$

8 0

3 years ago