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3 years ago

If an investor wants to combine a 5/8 acre parcel with a 3.9 acre parcel, how big will the parcel be, expressed as a decimal?

Mathematics

1 answer:

ipn [44]3 years ago

8 0

5/8 = 0.625

5/8 + 3.9 =
0.625 + 3.9 =
4.525 acre parcel <===

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Dress works in a factory that makes combs. He discovered that out of 500 combs, 34 of those are defective. If the factory produc

Bingel [31]

Answer: 175

Step-by-step explanation: First, a group of five hundred combs produces 34 defective combs. So if they make 2,500, you multiply how many times you add 500 to get to 2,500, which should be 5. Now multiply that to 34 and you get 175! To check it, divide: 34 divide by 150. Hope this helps!

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3 years ago

4p + 1 > −7 or 6p + 3 < 33?

aleksandr82 [10.1K]

P>-2 or p < 5 is your answer

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3 years ago

Solve the initial value problem where y′′+4y′−21 y=0, y(1)=1, y′(1)=0 . Use t as the independent variable.

igor_vitrenko [27]

Answer:

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

Step-by-step explanation:

y′′ + 4y′ − 21y = 0

The auxiliary equation is given by

m² + 4m - 21 = 0

We solve this using the quadratic formula. So

$m = \frac{-4 +/- \sqrt{4^{2} - 4 X 1 X (-21))} }{2 X 1}\\ = \frac{-4 +/- \sqrt{16 + 84} }{2}\\= \frac{-4 +/- \sqrt{100} }{2}\\= \frac{-4 +/- 10 }{2}\\= -2 +/- 5\\= -2 + 5 or -2 -5\\= 3 or -7$

So, the solution of the equation is

$y = Ae^{m_{1} t} + Be^{m_{2} t}$

where m₁ = 3 and m₂ = -7.

So,

$y = Ae^{3t} + Be^{-7t}$

Also,

$y' = 3Ae^{3t} - 7e^{-7t}$

Since y(1) = 1 and y'(1) = 0, we substitute them into the equations above. So,

$y(1) = Ae^{3X1} + Be^{-7X1}\\1 = Ae^{3} + Be^{-7}\\Ae^{3} + Be^{-7} = 1 (1)$

$y'(1) = 3Ae^{3X1} - 7Be^{-7X1}\\0 = 3Ae^{3} - 7Be^{-7}\\3Ae^{3} - 7Be^{-7} = 0 \\3Ae^{3} = 7Be^{-7}\\A = \frac{7}{3} Be^{-10}$

Substituting A into (1) above, we have

$\frac{7}{3}B e^{-10}e^{3} + Be^{-7} = 1 \\\frac{7}{3}B e^{-7} + Be^{-7} = 1\\\frac{10}{3}B e^{-7} = 1\\B = \frac{3}{10} e^{7}$

Substituting B into A, we have

$A = \frac{7}{3} \frac{3}{10} e^{7}e^{-10}\\A = \frac{7}{10} e^{-3}$

Substituting A and B into y, we have

$y = Ae^{3t} + Be^{-7t}\\y = \frac{7}{10} e^{-3}e^{3t} + \frac{3}{10} e^{7}e^{-7t}\\y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

So the solution to the differential equation is

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

6 0

3 years ago

According to the U.S. Bureau of Labor Statistics, 20% of all people 16 years of age or older do volunteer work. In this age grou

Murljashka [212]

Answer:

1. P(X≥35) = 0.0183

2. P(X≤21) = 0.0183

3. P(0.18<p<0.25) = 0.7915

Step-by-step explanation:

We have the proportion for women: pw=0.22, and the proportion for men: pm=0.19.

1. We have a sample of 140 woman and we have to calculate the probability of getting 35 or more who do volunteer work.

This is equivalent to a proportion of

$p=X/n=35/140=0.25$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.22*0.78}{300}}\\\\\\ \sigma_p=\sqrt{0.0006}=0.0239$

We calculate the z-score as:

$z=\dfrac{p-p_w}{\sigma_p}=\dfrac{0.25-0.22}{0.0239}=\dfrac{0.03}{0.0239}=0.8198$

Then, the probability of having 35 women or more who do volunteer work in this sample of 140 women is:

$P(X>35)=P(p>0.25)=P(z>2.0906)=0.0183$

2. We have to calculate the probability of having 21 or fewer women in the group who do volunteer work.

The proportion is now:

$p=X/n=21/140=0.15$

We can calculate then the z-score as:

$z=\dfrac{p-p_w}{\sigma_p}=\dfrac{0.15-0.2}{0.0239}=\dfrac{-0.05}{0.0239}=-2.0906$

Then, the probability of having 21 women or less who do volunteer work in this sample of 140 women is:

$P(X$

3. For the sample with men and women, we use the proportion for both, which is π=0.2.

The sample size is n=300.

Then, the standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.2*0.8}{300}}\\\\\\ \sigma_p=\sqrt{0.0005}=0.0231$

We can calculate the z-scores for p1=0.18 and p2=0.25:

$z_1=\dfrac{p_1-\pi}{\sigma_p}=\dfrac{0.18-0.2}{0.0231}=\dfrac{-0.02}{0.0231}=-0.8660\\\\\\z_2=\dfrac{p_2-\pi}{\sigma_p}=\dfrac{0.25-0.2}{0.0231}=\dfrac{0.05}{0.0231}=2.1651$

We can now calculate the probabilty of having a proportion within 0.18 and 0.25 as:

$P=P(0.18$

5 0

3 years ago

In a recent year, the weather was partly cloudy 2/5 of the days. Assuming there are 365 days in a year, how many days were partl

wariber [46]

365÷5=73 73×5=365
73×2=146 146÷2=73

146

7 0

3 years ago