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Pachacha [2.7K]
3 years ago
10

Josh makes $6916.67 per month. It is recommended to keep your housing budget between 25% 30% of your monthly income, how much ca

n Josh afford for rent?
Josh can afford rent between and $ per month (Round to the nearest cent)
Help me plz
Mathematics
1 answer:
Sphinxa [80]3 years ago
7 0

Take the monthly amount and times it by 25 30% and then you will get a small number. that number is his monthly income for his house

PS i dont have a cal. so i dont know the number

Step-by-step explanation:

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What is 12/7 please help and i will be very happy
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Answer:

12/7 is a improper fraction

Step-by-step explanation:

as a decimal it is 1.71428571429 and as a mixed number it is 1 5/7

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Find the area of the parallelogram. Please this is due today. Thank you.
scZoUnD [109]
894 mi^2

as 24 x 37.25 = 894
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What is the measure ??
alexandr402 [8]
Welcome

d the answer is 73 degrees
114 - 41 = 73
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3 years ago
The sodium content of a popular sports drink is listed as 200 mg in a 32-oz bottle. Analysis of 20 bottles indicates a sample me
julia-pushkina [17]

Answer:

The sample does not contradicts the manufacturer's claim.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 200 mg

Sample mean, \bar{x} = 211.5 mg

Sample size, n = 20

Alpha, α = 0.05

Sample standard deviation, s = 18.5 mg

a) First, we design the null and the alternate hypothesis  for a two tailed test

H_{0}: \mu = 200\\H_A: \mu \neq 200

We use Two-tailed t test to perform this hypothesis.

b) Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} } Putting all the values, we have

t_{stat} = \displaystyle\frac{211.5 - 200}{\frac{18.5}{\sqrt{20}} } = 2.7799

c) Now,

t_{critical} \text{ at 0.05 level of significance, 19 degree of freedom } = \pm 2.0930

Since, the calculated t-statistic does not lie in the range of the acceptance region(-2.0930,2.0930), we reject the null hypothesis.

Thus, the sample does not contradicts the manufacturer's claim.

d) P-value = 0.011934

Since the p-value is less than the significance level, we reject the null hypothesis and accept the alternate hypothesis.

Yes, both approach the critical value and the p-value approach gave the same results.

4 0
3 years ago
Suppose that surface σ is parameterized by r(u,v)=⟨ucos(3v),usin(3v),v⟩, 0≤u≤7 and 0≤v≤2π3 and f(x,y,z)=x2+y2+z2. Set up the sur
Bad White [126]

Looks like you have most of the details already, but you're missing one crucial piece.

\sigma is parameterized by

\vec r(u,v)=\langle u\cos3v,u\sin3v,v\rangle

for 0\le u\le7 and 0\le v\le\frac{2\pi}3, and a normal vector to this surface is

\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left\langle\sin3v,-\cos3v,3u\right\rangle

with norm

\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=\sqrt{\sin^23v+(-\cos3v)^2+(3u)^2}=\sqrt{9u^2+1}

So the integral of f(x,y,z)=x^2+y^2+z^2 is

\displaystyle\iint_\sigma f(x,y,z)\,\mathrm dA=\boxed{\int_0^{2\pi/3}\int_0^7(u^2+v^2)\sqrt{9u^2+1}\,\mathrm du\,\mathrm dv}

6 0
3 years ago
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