MATHEMATICS COLLEGE

A second number is 7 less than the first number. The third number is twice the first number. If the sum of the three numbers is 321, find the numbers

Answers

Answer 1

Answer:

Answer:

x=4

Step-by-step explanation:

hope it will help

First no.=x

2nd no.=2x

3rd no. =2x+6

x+2x+2x+6=26

5x+6=26

5×=26-6

5x =20.......divide both sides by 5

x=4

1st no.=x=4

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From data gathered in the period 2008−2012, the yearly value of U.S. exports can be modeled by the function E(x) = −228x3 + 2,252.8x2 − 6,098.5x + 11,425.8 where x is the number of years after 2008 and E(x) is the value of exports in billions of dollars. The yearly value of U.S. imports can be modeled by the function I(x) = −400.4x3 + 3,954.4x2 − 11,128.8x + 17,749.6 where x is the number of years after 2008 and I(x) is the value of imports in billions of dollars. Estimate the total value the U.S. imported and exported in 2011. Round to the nearest whole number. About billion dollars were imported and exported by the U.S. in 2011.
Imari spent half her weekly allowance playing mini-golf. To earn more money, her parents let her wash the car for $4. If she ended up with $12 how much is her weekly allowance? (Prove Your Answer)
Linda found that the cost to get a swimming pool installed in her backyard is a linear function of the pool's area. A swimming pool with an area of 1,000 square feet can be installed for $50,000, whereas the installation of an 800 square foot swimming pool costs $35,000. Select the correct graph that models the given relationship.
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Jeremy sprinted for 123 seconds and rested. Then he sprinted for 157 seconds, rested, and sprinted again for 195 seconds. Estimate the combined time he sprinted by rounding to the nearest ten and then adding the rounded numbers.

you save $15000. you place one third in a savings account earning 4.6% apr compounded annually. you then invest one quarter of the remaining balance in a 3 year u.s treasury bond earning 5.2 apr compounded annually and the rest in a stock plan. your stock plan increases in value 3% the first year decrease 8%in value the second year and increases 6% in value in the third year. what are the balances for each account by the end of the third year and the total gain in your original saved amount

Answers

Given initial savings are 15,000 dollars.

It says to place one-third in an account with 4.6% APR compounded annually. So we invest $5,000 in this account.

Amount after 3 years = 15000(1+0.046)³ = 17,166.68 dollars.

Remaining Balance = 15,000 - 5,000 = 10,000 dollars.

It says to place one-quarter of the remaining balance into 3-year bond with 5.2% APR compounded annually. So we invest $2,500 in this bond.

Bond's value after 3 years = 2500(1+0.052)³ = 2,910.63 dollars.

Remaining Balance = 10,000 - 2,500 = 7,500 dollars.

It says to invest rest in a stock that increases in value 3% the first year; decrease 8% in value the second year; and increases 6% in value in the third year.

Stock's value after 3 years = 7500×(103%)×(92%)×(106%) = 7500 x 1.03 x 0.92 x 1.06 = 7,533.42 dollars.

Total amount after 3 years = Saving Account + Bond value + Stock value = 17,166.68 + 2,910.63 + 7,533.42 = 27,610.73 dollars.

Total Gain in the original savings = 27,610.73 - 15,000 = 12,610.73 dollars

100 dollars a day because when u do the math u see that the division is th key

Use a triple integral to find the volume of the solid enclosed by the paraboloid x=6y2 6z2 and the plane x=6.

Answers

I assume you mean the paraboloid $x=6y^2+6z^2$ . Denote the space between this surface and the plane $x=6$ by the symbol $D$ .

The volume is then

$\displaystyle\iiint_D\mathrm dV=\int_(-1)^1\int_(-1)^1\int_(6y^2+6z^2)^6\mathrm dx\,\mathrm dy\,\mathrm dz=8$

Can you please give me a short brief answer

Answers

Answer:

120 yards 1.

Step-by-step explanation:

In the shape, it is a rectangle with a triangle cut out from it. What is the area of the shape:

Area of rectangle = length × width

12 × 15 = 180 yards²

Area of triangle = ½ × base × height

½ × 10 × 12 = 60 yards²

Now subtract the area of triangle from the area of rectangle:

180 - 60 = 120 yards ²

Please help me out!!!

Answers

Answer:

Shifts 4 units down ---> $g(x)=2x-10$

Stretches f(x) by a factor of 4 away from x-axis---> $g(x)=8x-6$

Shifts f(x) 4 units right---> $g(x)=2x-14$

Compress f(x) by a factor of 1/4 toward the y-axis ---> $g(x)=1/2x-3/2$

Step-by-step explanation:

We are given $f(x)=2x-6$

We need to match the transformations.

1) shifts f(x) 4 units down.

When function f(x) shifts k units down the new function becomes f(x)-k

In our case

$g(x)=2x-6-4\ng(x)=2x-10$

So, Shifts 4 units down ---> $g(x)=2x-10$

2) Stretches f(x) by a factor of 4 away from x-axis

When function f(x) is stretched by a factor of b away from x-axis the new function becomes f(bx)

$g(x)=2(4x)-6\ng(x)=8x-6$

So, Stretches f(x) by a factor of 4 away from x-axis---> $g(x)=8x-6$

3) Shifts f(x) 4 units right

When function f(x) shifts h units right the new function becomes f(x-h)

$g(x)=2(x-4)-6\ng(x)=2x-8-6\ng(x)=2x-14$

So, Shifts f(x) 4 units right---> $g(x)=2x-14$

4) Compress f(x) by a factor of 1/4 toward the y-axis

When function f(x) is compressed by h factor of a toward the y-axis the new function becomes h.f(x)

$g(x)=1/4(2x-6)\ng(x)=1/2x-3/2$

Compress f(x) by a factor of 1/4 toward the y-axis ---> $g(x)=1/2x-3/2$

(Option Not given)

(If we compress f(x) by a factor of 4 towards y-axis we get g(x)=8x-24)

"The municipal transportation authority determined that 58% of all drivers were speeding along a busy street. In an attempt to reduce this percentage, the city put up electronic speed monitors so that drivers would be warned if they were driving over the speed limit. A follow-up study is now planned to see if the speed monitors were effective. The null and alternative hypotheses of the test are H0 : π = 0.58 versus Ha : π < 0.58. It is planned to use a sample of 150 drivers taken at random times of the week and the test will be conducted at the 5% significance level. (a) What is the most number of drivers in the sample that can be speeding and still have them conclude that the alternative hypothesis is true? (b) Suppose the true value of π is 0.52. What is the power of the test? (c) How could the researchers modify the test in order to increase its power without increasing the probability of a Type I Error?"

Answers

Answer:

a) X=77 drivers

b) Power of the test = 0.404

c) Increasing the sample size.

Step-by-step explanation:

This is a hypothesis test of proportions. As the claim is that the speed monitors were effective in reducing the speeding, this is a left-tail test.

For a left-tail test at a 5% significance level, we have a critical value of z that is zc=-1.645. This value is the limit of the rejection region. That means that if the test statistic z is smaller than zc=-1.645, the null hypothesis is rejected.

The proportion that would have a test statistic equal to this critical value can be expressed as:

$p_c=\pi+z_c\cdot\sigma_p$

The standard error of the proportion is:

$\sigma_p=\sqrt{(\pi(1-\pi))/(n)}=\sqrt{(0.58*0.42)/(150)}\n\n\n \sigma_p=√(0.001624)=0.04$

Then, the proportion is:

$p_c=\pi+z_c\cdot\sigma_p=0.58-1.645*0.04=0.58-0.0658=0.5142$

This proportion, with a sample size of n=150, correspond to

$x=n\cdot p=150\cdot0.5142=77.13\approx 77$

The power of the test is the probability of correctly rejecting the null hypothesis.

The true proportion is 0.52, but we don't know at the time of the test, so the critical value to make a decision about rejecting the null hypothesis is still zc=-1.645 corresponding to a critical proportion of 0.51.

Then, we can say that the probability of rejecting the null hypothesis is still the probability of getting a sample of size n=150 with a proportion of 0.51 or smaller, but within a population with a proportion of 0.52.

The standard error has to be re-calculated for the new true proportion:

$\sigma_p=\sqrt{(\pi(1-\pi))/(n)}=\sqrt{(0.52*0.48)/(150)}\n\n\n \sigma_p=√(0.001664)=0.041$

Then, we calculate the z-value for this proportion with the true proportion:

$z=(p-\pi')/(\sigma_p)=(0.51-0.52)/(0.041)=(-0.01)/(0.041)=-0.244$

The probability of getting a sample of size n=150 with a proportion of 0.51 or lower is:

$P(p<0.51)=P(z<-0.244)=0.404$

Then, the power of the test is β=0.404.

The only variable left to change in the test in order to increase the power of the test is the sample size, as the significance level can not be changed (it is related to the probability of a Type I error).

It the sample size is increased, the standard error of the proprotion decreases. As the standard error tends to zero, the critical proportion tend to 0.58, as we can see in its equation:

$\lim_(\sigma_p \to 0) p_c=\pi+ \lim_(\sigma_p \to 0)(z_c\cdot\sigma_p)=\pi=0.58$