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Algebraic expression for three more than a number is twenty less than three times the same number write equation form

Answers

Answer 1

Answer:

Answer:

x+3 = 3x-20

Step-by-step explanation:

Answer 2

Answer:

3+x = 20 - 3x

More than = adding

is is equal to

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The equation of a parabola is given.y=−14x2+4x−19

What are the coordinates of the vertex of the parabola?

Enter your answer in the boxes.

Answers

Answer:

vertex ( $(1)/(7)$ , $(131)/(7)$ ).

Step-by-step explanation:

Given : The equation of a parabola is given.

y=−14x²+4x−19

To find : What are the coordinates of the vertex of the parabola.

Solution : We have given that

y = −14x²+4x−19

we will be "completing the square" .

Factor out -14 to make leading coefficient 1

y = -14 ( $x^(2) -(2x)/(7) +(19)/(14)$ )

Add and subtract $((-1)/(7)) ^(2)$

y = -14 ( $x^(2) -(2x)/(7) +(19)/(14)+(-1)/(7)) ^(2) - ((-1)/(7))^(2))$ .

Complete the square

y = -14 ( $(x -(1)/(7)) ^(2) +(19)/(14)- ((-1)/(7))^(2))$ .

y = -14 ( $(x-(1)/(7)) ^(2) -(131)/(7)$

Standard form of parabola vertex form y = a(x - h)²+ k,

Where, ( h, k) are vertex

On comparing a = -14 , h= $(1)/(7)$ , k = $(131)/(7)$

Therefore, vertex ( $(1)/(7)$ , $(131)/(7)$ ).

The coordinates are 8, -3.

TV advertising agencies face growing challenges in reaching audience members because viewing TV programs via digital streaming is increasingly popular. the Harris poll reported on November 13, 2012, that 53% of 2343 American adults surveyed said they have watched digitally streamed TV programming on some type of device.a. calculate and interpret a confidence interval at the 99% confidence level for the proportion of all adult americans who have watched streamed programming.b. what sample size would be required for the width of a 99% CI to be at most .05 irrespective of the value of p?

Answers

Answer:

a) The 99% confidence interval would be given (0.503;0.557).

We are 99% confident that this interval contains the true population proportion.

b) $n=(0.53(1-0.53))/(((0.05)/(2.58))^2)=663.2$

And rounded up we have that n=664

Step-by-step explanation:

Data given and notation

n=2343 represent the random sample taken

X represent the people that they have watched digitally streamed TV programming on some type of device

$\hat p=0.53$ estimated proportion of people that they have watched digitally streamed TV programming on some type of device

$\alpha=0.01$ represent the significance level

Confidence =0.99 or 99%

z would represent the statistic for the confidence interval

p= population proportion of people that they have watched digitally streamed TV programming on some type of device

The population proportion present the following distribution:

$p \sim N (p, \sqrt{(p(1-p))/(n)}$

Part a) Confidence interval

The confidence interval would be given by this formula

$\hat p \pm z_(\alpha/2) \sqrt{(\hat p(1-\hat p))/(n)}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_(\alpha/2)=2.58$

And replacing into the confidence interval formula we got:

$0.53 - 2.58 \sqrt{(0.53(1-0.53))/(2343)}=0.503$

$0.53 + 2.58 \sqrt{(0.53(1-0.53))/(2343)}=0.557$

And the 99% confidence interval would be given (0.503;0.557).

We are 99% confident that this interval contains the true population proportion.

Part b) What sample size would be required for the width of a 99% CI to be at most 0.05 irrespective of the value of p??

The margin of error for the proportion interval is given by this formula:

$ME=z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}$ (a)

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\hat p (1-\hat p))/(((ME)/(z))^2)$ (b)

And replacing into equation (b) the values from part a we got:

$n=(0.53(1-0.53))/(((0.05)/(2.58))^2)=663.2$

And rounded up we have that n=664

A painter can finish painting a house in 3 hour

Answers

Answer:

cool

Step-by-step explanation:

An owner of a key rings manufacturing company found that the profit earned (in thousands of dollars) per day by selling n number of key rings is given by , where n is the number of key rings in thousands. Find the number of key rings sold on a particular day when the total profit is $5,000.

Answers

Answer:

The number of key rings sold on a particular day when the total profit is $5,000 is 4,000 rings.

Step-by-step explanation:

The question is incomplete.

An owner of a key rings manufacturing company found that the profit earned (in thousands of dollars) per day by selling n number of key rings is given by

$P=n^2-2n-3$

where n is the number of key rings in thousands.

Find the number of key rings sold on a particular day when the total profit is $5,000.

We have the profit defined by a quadratic function.

We have to calculate n, for which the profit is $5,000.

$P=n^2-2n-3=5\n\nn^2-2n-8=0$

We have to calculate the roots of the polynomial we use the quadratic equation:

$n=(-b\pm√(b^2-4ac))/(2a)\n\nn= (-2\pm√(4-4*1*(-8)))/(2)= (-2\pm√(4-32))/(2) = (-2\pm√(36))/(2) =(-2\pm6)/(2) \n\nn_1=(-2-6)/2=-8/2=-4\n\nn_2=(-2+6)/2=4/2=2$

n1 is not valid, as the amount of rings sold can not be negative.

Then, the solution is n=4 or 4,000 rings sold.

A bag contains some number of marbles. It is known that 20 of them are red. When 15 marbles are drawn, without replacement, we get 6 red. Assuming E(X)=6 red, what is the total number of marbles in the bag?

Answers

Answer:

The total number of marbles in the bag is 50.

Step-by-step explanation:

Here, we have n trials, without replacement. So the hypergeometric distribution is used.

The mean of the hypergeometric distribution is:

$E(X) = (n*k)/(N)$

In which n is the number of items in the sample, k is the number of items in the population that are classified a success and N is the size of the population.

15 marbles are drawn:

This means that $n = 15$