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LekaFEV [45]
3 years ago
10

Sammie's monthly take-home pay is $3600, and his monthly rent is $900. If both his monthly take-home pay and his rent increase b

y $250, what percentage of Sammie's take-home pay will be used to pay rent?
Mathematics
2 answers:
4vir4ik [10]3 years ago
8 0

The original salary = $ 3,600

The increased salary will be = $ 3,600 + $ 250 = $ 3,850

The original rent = $ 900

The increased rent will be = $ 900 + $ 250

Since, both rent and salary increased by $ 250.

Now, the percentage of Sammie' salary paid as rent will be calculated as -

Percentage of salary paid as rent = \frac{Increased Rent}{Increased salary} × 100

Percentage of salary paid as rent = \frac{$ 1,150}{$ 3,850} × 100

<u>Percentage of salary paid as rent = 29.9%</u>

USPshnik [31]3 years ago
4 0
WELL,
9<span>00/3600=.25
=25%</span>
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Today i will walk 3 miles per hour to school, which is 1.5 miles away how many hours would this trip take
GaryK [48]

Answer: 30 minutes so 1/2 hour

Step-by-step explanation:

Because 3 miles = 1 hour

1.5 times 2 = 3

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4 0
3 years ago
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An article presents voltage measurements for a sample of 66 industrial networks in Estonia. Assume the rated voltage for these n
kramer

Answer:

We accept the null hypothesis and conclude that voltage for these networks is 232 V.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 232 V

Sample mean, \bar{x} = 231.5 V

Sample size, n = 66

Sample standard deviation, s = 2.19 V

Alpha, α = 0.05

First, we design the null and the alternate hypothesis

H_{0}: \mu = 232\\H_A: \mu \neq 232

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

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n-1}} } Putting all the values, we have

t_{stat} = \displaystyle\frac{231.5- 232}{\frac{2.19}{\sqrt{66}} } = -1.8548

Now,

t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = \pm 1.9971

Since,              

|t_{stat}| > |t_{critical}|

We accept the null hypothesis and conclude that voltage for these networks is 232 V.

4 0
3 years ago
1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves
erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx

\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx

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When x = 0, we have y = 2, so we solve for the constant C :

\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)

Then the particular solution to the DE is

\ln|y| = x - \ln|x+1| + \ln(2)

We can go on to solve explicitly for y in terms of x :

e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}

(b) The curves y = x² and y = 2x - x² intersect for

x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1

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\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}

The area of this region is

\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}

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I'm sure this is the answer

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jarptica [38.1K]
You must add another 600ml acid. Good luck and pls mark as brainliest if I helped
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