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2 years ago

In a year group of 108 students, 60 liked football,

Mathematics

1 answer:

kap26 [50]2 years ago

8 0

Answer:

you know what to do that is the soup for the next couple of weeks and see how much is a ps4 controller or an official at the lot of the boy's

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Adiya said that the first step to solving the quadratic equation x2 + 6 = 20x by completing the square was to divide 6 by 2, squ

kicyunya [14]

Adiya’s method is not correct. To form a perfect square trinomial, the constant must be isolated on one side of the equation. Also, the coefficient of the term with an exponent of 1 on the variable is used to find the constant in the perfect square trinomial. Adiya should first get the 20x term on the same side of the equation as x2. Then she would divide 20 by 2, square it, and add 100 to both sides.

6 0

2 years ago

Read 2 more answers

Jenna wants to open a salon and must choose a location. At Location A, she will pay $1200 per month and charge $45 per haircut.

anzhelika [568]

Profit=revenue-cost

pA(m,h)=45h-1200m

pB(m,h)=60h-1800m

Profit is the same when pA=pB so

60h-1800m=45h-1200m

15h=600m

h=40m

So 40 haircuts per month at either location results in the same profit of $600 per month.

5 0

2 years ago

James folds a piece of paper in half several times,each time unfolding the paper to count how many equal parts he sees. After fo

Snezhnost [94]

Answer:

There will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

The needed function is $y = 2 ^n$

Step-by-step explanation:

Let us assume the piece of paper James decides to fold is a SQUARE.

Now, let us assume:

n : the number of times the paper is folded.

y : The number of parts obtained after folds.

Now, if the paper if folded ONCE ⇒ n = 1

Also, when the pap er is folded once, the parts obtained are TWO equal parts.

⇒ for n = 1 , y = 2 ..... (1)

Similarly, if the paper if folded TWICE ⇒ n = 2

Also, when the paper is folded twice, the parts obtained are FOUR equal parts.

⇒ for n = 2 , y = 4 ..... (2)

⇒ $y = 2^2 = 2^n$

Continuing the same way, if the paper is folded SEVEN times ⇒ n = 7

So, $y = 2^ n = 2^7 = 128$

⇒ There are total 128 equal parts.

Lastly, if the paper is folded ELEVEN times ⇒ n = 11

So, $y = 2^ n = 2^{11} = 2048$

⇒ There are total 2048 equal parts.

Hence, there will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

And the needed function is $y = 2 ^n$

8 0

2 years ago

Kiran has 27 nickels and quarters in his pocket, worth a total of $2.75.

Montano1993 [528]

The system of equations is

n + d = 27

n + 5d = 55

Step-by-step explanation:

The given is:

Kiran has 27 nickels and quarters in his pocket
They worth a total of $2.75
We need to write a system of equations to represent the relationships between the number of nickels n, the number of quarters d, and the dollar amount in this situation

∵ The number of nickles is n

∵ The number of quarters is d

∵ There are 27 nickles and quarters

∴ n + d = 27 ⇒ (1)

∵ 1 nickel = 5 cents

∵ 1 quarter = 25 cents

- Multiply n by 5 and d by 25 to find their value

∴ They worth = 5n + 25d

∵ They worth a total of $2.75

- Chang the dollar to cent

∵ 1 dollar = 100 cents

∴ $2.75 = 2.75 × 100 = 275 cents

- Equate their value by 275

∴ 5n + 25d = 275

- Simplify it by divide each term by 5

∴ n + 5d = 55 ⇒ (2)

The system of equations is

n + d = 27

n + 5d = 55

Learn more:

You can learn more about the system of equations in brainly.com/question/2115716

#LearnwithBrainly

5 0

2 years ago

The Fortune 500 is comprised of Fortune magazine's annual list of the finite population of the top 500 companies in the U.S. ran

Serjik [45]

Answer:

The standard error of the proportion is 0.0367.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the standard error is $s = \sqrt{\frac{p(1-p)}{n}}$

In this question:

$p = 0.16, n = 100$

$s = \sqrt{\frac{0.16*0.84}{100}} = 0.0367$

The standard error of the proportion is 0.0367.

4 0

2 years ago